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3.3: Matukio ya kujitegemea na ya kipekee

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    Huru na ya kipekee haimaanishi kitu kimoja.

    Matukio ya kujitegemea

    Matukio mawili yanajitegemea ikiwa yafuatayo ni ya kweli:

    • \(P(\text{A|B}) = P(\text{A})\)
    • \(P(\text{B|A}) = P(\text{B})\)
    • \(P(\text{A AND B}) = P(\text{A})P(\text{B})\)

    Matukio mawili\(\text{A}\) na\(\text{B}\) ni huru kama ujuzi kwamba moja ilitokea hauathiri nafasi nyingine hutokea. Kwa mfano, matokeo ya majukumu mawili ya kufa kwa haki ni matukio ya kujitegemea. Matokeo ya roll ya kwanza haina mabadiliko ya uwezekano wa matokeo ya roll ya pili. Kuonyesha matukio mawili ni huru, lazima uonyeshe moja tu ya masharti hapo juu. Ikiwa matukio mawili hayategemea, basi tunasema kuwa wanategemea.

    Sampuli ya idadi ya watu

    Sampuli inaweza kufanyika kwa uingizwaji au bila uingizwaji (Kielelezo\(\PageIndex{1}\)):

    • Kwa uingizwaji: Ikiwa kila mwanachama wa idadi ya watu hubadilishwa baada ya kuchukuliwa, basi mwanachama huyo ana uwezekano wa kuchaguliwa zaidi ya mara moja. Wakati sampuli imefanywa na uingizwaji, basi matukio yanachukuliwa kuwa huru, maana ya matokeo ya pick ya kwanza haitabadilika uwezekano wa kuchukua pili.
    • Bila uingizwaji: Wakati sampuli inafanywa bila uingizwaji, kila mwanachama wa idadi ya watu anaweza kuchaguliwa mara moja tu. Katika kesi hii, uwezekano wa kuchukua pili huathiriwa na matokeo ya pick ya kwanza. Matukio yanachukuliwa kuwa tegemezi au si huru.
    Kielelezo\(\PageIndex{1}\): uwakilishi Visual ya mchakato sampuli. Ikiwa vitu vya sampuli vinabadilishwa baada ya kila tukio la sampuli, basi hii ni “sampuli na uingizwaji” ikiwa sio, basi ni “sampuli bila uingizwaji”. Picha imetumiwa kwa ruhusa (CC BY-SA 4.0; Dan Kernler).

    Kama haijulikani kama\(\text{A}\) na\(\text{B}\) ni huru au tegemezi, kudhani wao ni tegemezi mpaka unaweza kuonyesha vinginevyo.

    Mfano\(\PageIndex{1}\): Sampling with and without replacement

    Una haki, vizuri shuffled staha ya kadi 52. Inajumuisha suti nne. Suti ni vilabu, almasi, mioyo na spades. Kuna kadi 13 katika kila suti yenye 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\(\text{J}\) (jack), (malkia),\(\text{K}\) (mfalme) wa suti hiyo.\(\text{Q}\)

    a. sampuli na uingizwaji:

    Tuseme unachukua kadi tatu na uingizwaji. kadi ya kwanza wewe kuchukua nje ya 52 kadi ni\(\text{Q}\) ya spades. Kuweka kadi hii nyuma, reshuffle kadi na kuchukua kadi ya pili kutoka 52-kadi ya staha. Ni kumi ya vilabu. Kuweka kadi hii nyuma, reshuffle kadi na kuchukua kadi ya tatu kutoka 52-kadi ya staha. Wakati huu, kadi ni\(\text{Q}\) ya spades tena. Tar yako ni {\(\text{Q}\)ya spades, kumi ya vilabu,\(\text{Q}\) ya spades}. Umechukua\(\text{Q}\) ya spades mara mbili. Kuchukua kila kadi kutoka 52-kadi ya staha.

    b Sampuli bila uingizwaji:

    Tuseme wewe kuchukua kadi tatu bila uingizwaji. kadi ya kwanza wewe kuchukua nje ya 52 kadi ni\(\text{K}\) ya mioyo. Kuweka kadi hii kando na kuchukua kadi ya pili kutoka 51 kadi iliyobaki katika staha. Ni tatu za almasi. Kuweka kadi hii kando na kuchukua kadi ya tatu kutoka iliyobaki 50 kadi katika staha. Kadi ya tatu ni\(\text{J}\) ya spades. Tar yako ni {\(\text{K}\)ya mioyo, tatu ya almasi,\(\text{J}\) ya spades}. Kwa sababu umechukua kadi bila uingizwaji, huwezi kuchukua kadi hiyo mara mbili.

    Zoezi\(\PageIndex{1}\)

    Una haki, vizuri shuffled staha ya kadi 52. Inajumuisha suti nne. Suti ni vilabu, almasi, mioyo na spades. Kuna kadi 13 katika kila suti yenye 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\(\text{J}\) (jack), (malkia),\(\text{K}\) (mfalme) wa suti hiyo.\(\text{Q}\) Kadi tatu ni ilichukua kwa random.

    1. Tuseme unajua kwamba kadi ilichukua ni\(\text{Q}\) ya spades,\(\text{K}\) ya mioyo na\(\text{Q}\) ya spades. Je, unaweza kuamua kama sampuli ilikuwa na au bila badala?
    2. Tuseme unajua kwamba kadi ilichukua ni\(\text{Q}\) ya spades,\(\text{K}\) ya mioyo, na\(\text{J}\) ya spades. Je, unaweza kuamua kama sampuli ilikuwa na au bila badala?
    Jibu

    Pamoja na uingizwaji

    Jibu b

    Hapana

    Mfano\(\PageIndex{2}\)

    Una haki, vizuri shuffled staha ya kadi 52. Inajumuisha suti nne. Suti ni vilabu, almasi, mioyo, na spades. Kuna kadi 13 katika kila suti yenye 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\(\text{J}\) (jack),\(\text{Q}\) (malkia), na\(\text{K}\) (mfalme) wa suti hiyo. \(\text{S} =\)spades,\(\text{H} =\) Hearts,\(\text{D} =\) Diamonds,\(\text{C} =\) Vilabu.

    1. Tuseme wewe kuchukua kadi nne, lakini si kuweka kadi yoyote nyuma katika staha. Kadi yako ni\(\text{QS}, 1\text{D}, 1\text{C}, \text{QD}\).
    2. Tuseme wewe kuchukua kadi nne na kuweka kila kadi nyuma kabla ya kuchukua kadi ya. Kadi yako ni\(\text{KH}, 7\text{D}, 6\text{D}, \text{KH}\).

    Ni ipi kati ya. au b. ulifanya sampuli na uingizwaji na ni ipi uliyofanya sampuli bila uingizwaji?

    Jibu

    Bila uingizwaji

    Jibu b

    Pamoja na uingizwaji

    Zoezi\(\PageIndex{2}\)

    Una haki, vizuri shuffled staha ya kadi 52. Inajumuisha suti nne. Suti ni vilabu, almasi, mioyo, na spades. Kuna kadi 13 katika kila suti yenye 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\(\text{J}\) (jack),\(\text{Q}\) (malkia), na\(\text{K}\) (mfalme) wa suti hiyo. \(\text{S} =\)spades,\(\text{H} =\) Hearts,\(\text{D} =\) Diamonds,\(\text{C} =\) Vilabu. Tuseme kwamba sampuli kadi nne bila uingizwaji. Ni ipi kati ya matokeo yafuatayo yanawezekana? Jibu swali moja kwa sampuli na uingizwaji.

    1. \(\text{QS}, 1\text{D}, 1\text{C}, \text{QD}\)
    2. \(\text{KH}, 7\text{D}, 6\text{D}, \text{KH}\)
    3. \(\text{QS}, 7\text{D}, 6\text{D}, \text{KS}\)
    Jibu - bila uingizwaji

    Inawezekana; b. haiwezekani, c. inawezekana

    Jibu - na uingizwaji

    Inawezekana; c. inawezekana, c. inawezekana

    Matukio ya kipekee

    \(\text{A}\)na\(\text{B}\) ni matukio ya kipekee ikiwa hayawezi kutokea kwa wakati mmoja. Hii ina maana kwamba\(\text{A}\) na\(\text{B}\) wala kushiriki matokeo yoyote na\(P(\text{A AND B}) = 0\).

    Kwa mfano, tuseme nafasi ya sampuli

    \[S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \nonumber\]

    Hebu\(\text{A} = \{1, 2, 3, 4, 5\}, \text{B} = \{4, 5, 6, 7, 8\}\), na\(\text{C} = \{7, 9\}\). \(\text{A AND B} = \{4, 5\}\).

    \[P(\text{A AND B}) = \dfrac{2}{10} \nonumber\]

    na si sawa na sifuri. Kwa hiyo,\(\text{A}\) na\(\text{B}\) si pande kipekee. \(\text{A}\)na\(\text{C}\) hawana idadi yoyote kwa kawaida hivyo\(P(\text{A AND C}) = 0\). Kwa hiyo,\(\text{A}\) na\(\text{C}\) ni pande kipekee.

    Kama haijulikani kama\(\text{A}\) na\(\text{B}\) ni pande kipekee, kudhani wao si mpaka unaweza kuonyesha vinginevyo. Mifano zifuatazo zinaonyesha ufafanuzi na masharti haya.

    Mfano\(\PageIndex{3}\)

    Flip sarafu mbili za haki.

    Nafasi ya sampuli ni\(\{HH, HT, TH, TT\}\) wapi\(T =\) mikia na\(H =\) vichwa. Matokeo ni\(HH,HT, TH\), na\(TT\). Matokeo\(HT\) na\(TH\) ni tofauti. \(HT\)Njia ambazo sarafu ya kwanza ilionyesha vichwa na sarafu ya pili ilionyesha mikia. \(TH\)Njia ambazo sarafu ya kwanza ilionyesha mikia na sarafu ya pili ilionyesha vichwa.

    • Hebu tukio\(\text{A} =\) la kupata mkia mmoja zaidi. (Kwa mkia mmoja zaidi ina maana ya sifuri au mkia mmoja.) Kisha\(\text{A}\) inaweza kuandikwa kama\(\{HH, HT, TH\}\). Matokeo\(HH\) inaonyesha mikia ya sifuri. \(HT\)na\(TH\) kila kuonyesha mkia mmoja.
    • Hebu tukio\(\text{B} =\) la kupata mikia yote. \(\text{B}\)inaweza kuandikwa kama\(\{TT\}\). \(\text{B}\)ni inayosaidia\(\text{A}\), hivyo\(\text{B} = \text{A′}\). Pia,\(P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1\).
    • probabilities kwa\(\text{A}\) na kwa\(\text{B}\) ni\(P(\text{A}) = \dfrac{3}{4}\) na\(P(\text{B}) = \dfrac{1}{4}\).
    • Hebu tukio\(\text{C} =\) la kupata vichwa vyote. \(\text{C} = \{HH\}\). Tangu\(\text{B} = \{TT\}\),\(P(\text{B AND C}) = 0\). \(\text{B}\)na Care pande kipekee. \(\text{B}\)na\(\text{C}\) msiwe na viungo sawa kwa sababu huwezi kuwa na mikia yote na vichwa vyote kwa wakati mmoja.)
    • Hebu\(\text{D} =\) tukio la kupata mkia zaidi ya moja. \(\text{D} = \{TT\}\). \(P(\text{D}) = \dfrac{1}{4}\)
    • Hebu\(\text{E} =\) tukio la kupata kichwa kwenye roll ya kwanza. (Hii ina maana unaweza kupata kichwa au mkia kwenye roll ya pili.) \(\text{E} = \{HT, HH\}\). \(P(\text{E}) = \dfrac{2}{4}\)
    • Pata uwezekano wa kupata angalau moja (moja au mbili) mkia katika flips mbili. Hebu\(\text{F} =\) tukio la kupata angalau mkia mmoja katika flips mbili. \(\text{F} = \{HT, TH, TT\}\). \(P(\text{F}) = \dfrac{3}{4}\)

    Zoezi\(\PageIndex{3}\)

    Chora kadi mbili kutoka staha ya kiwango cha 52-kadi na uingizwaji. Pata uwezekano wa kupata angalau kadi moja nyeusi.

    Jibu

    Nafasi ya sampuli ya kuchora kadi mbili na uingizwaji kutoka kwenye staha ya kawaida ya kadi ya 52 kwa heshima na rangi ni\(\{BB, BR, RB, RR\}\).

    Tukio\(A =\) Kupata angalau kadi nyeusi\(= \{BB, BR, RB\}\)

    \(P(\text{A}) = \dfrac{3}{4} = 0.75\)

    Mfano\(\PageIndex{4}\)

    Flip sarafu mbili za haki. Pata uwezekano wa matukio.

    1. Hebu tukio\(\text{F} =\) la kupata mkia mmoja (sifuri au mkia mmoja).
    2. Hebu tukio\(\text{G} =\) la kupata nyuso mbili ambazo ni sawa.
    3. Hebu tukio\(\text{H} =\) la kupata kichwa kwenye flip ya kwanza ikifuatiwa na kichwa au mkia kwenye flip ya pili.
    4. Je\(\text{F}\), na\(\text{G}\) pande kipekee?
    5. Hebu tukio\(\text{J} =\) la kupata mikia yote. Je\(\text{J}\), na\(\text{H}\) pande kipekee?

    Suluhisho

    Angalia nafasi ya sampuli katika Mfano\(\PageIndex{3}\).

    1. Zero (0) au moja (1) mikia hutokea wakati matokeo\(HH, TH, HT\) yanavyoonekana. \(P(\text{F}) = \dfrac{3}{4}\)
    2. Nyuso mbili ni sawa kama\(HH\) au\(TT\) show up. \(P(\text{G}) = \dfrac{2}{4}\)
    3. Kichwa kwenye flip ya kwanza ikifuatiwa na kichwa au mkia kwenye flip ya pili hutokea wakati\(HH\) au\(HT\) kuonyesha. \(P(\text{H}) = \dfrac{2}{4}\)
    4. \(\text{F}\)na\(\text{G}\) kushiriki\(HH\) hivyo\(P(\text{F AND G})\) si sawa na sifuri (0). \(\text{F}\)na\(\text{G}\) si pande kipekee.
    5. Kupata mikia yote hutokea wakati mikia inavyoonekana kwenye sarafu zote mbili (\(TT\)). \(\text{H}\)'s matokeo ni\(HH\) na\(HT\).

    \(\text{J}\)na\(\text{H}\) kuwa na kitu sawa hivyo\(P(\text{J AND H}) = 0\). \(\text{J}\)na\(\text{H}\) ni pande kipekee.

    Zoezi\(\PageIndex{4}\)

    Sanduku lina mipira miwili, nyeupe moja na nyekundu moja. Sisi kuchagua mpira mmoja, kuiweka nyuma katika sanduku, na kuchagua mpira wa pili (sampuli na uingizwaji). Pata uwezekano wa matukio yafuatayo:

    1. Hebu tukio\(\text{F} =\) la kupata mpira mweupe mara mbili.
    2. Hebu tukio\(\text{G} =\) la kupata mipira miwili ya rangi tofauti.
    3. Hebu tukio\(\text{H} =\) la kupata nyeupe kwenye pick ya kwanza.
    4. Je\(\text{F}\), na\(\text{G}\) pande kipekee?
    5. Je\(\text{G}\), na\(\text{H}\) pande kipekee?

    Jibu

    1. \(P(\text{F}) = \dfrac{1}{4}\)
    2. \(P(\text{G}) = \dfrac{1}{2}\)
    3. \(P(\text{H}) = \dfrac{1}{2}\)
    4. Ndio
    5. Hapana

    Mfano\(\PageIndex{5}\)

    Roll haki moja, sita upande mmoja kufa. Nafasi ya sampuli ni {1, 2, 3, 4, 5, 6}. Hebu tukio uso\(\text{A} =\) ni isiyo ya kawaida. Kisha\(\text{A} = \{1, 3, 5\}\). Hebu tukio\(\text{B} =\) uso ni hata. Kisha\(\text{B} = \{2, 4, 6\}\).

    • Kupata inayosaidia ya\(\text{A}\),\(\text{A′}\). Msaidizi wa\(\text{A}\),\(\text{A′}\), ni\(\text{B}\) kwa sababu\(\text{A}\) na\(\text{B}\) pamoja hufanya nafasi ya sampuli. \(P(\text{A}) + P(\text{B}) = P(\text{A}) + P(\text{A′}) = 1\). pia,\(P(\text{A}) = \dfrac{3}{6}\) na\(P(\text{B}) = \dfrac{3}{6}\).
    • Hebu tukio\(\text{C} =\) isiyo ya kawaida nyuso kubwa kuliko mbili. Kisha\(\text{C} = \{3, 5\}\). Hebu tukio\(\text{D} =\) zote hata nyuso ndogo kuliko tano. Kisha\(\text{D} = \{2, 4\}\). \(P(\text{C AND D}) = 0\)kwa sababu huwezi kuwa na uso usio wa kawaida na hata kwa wakati mmoja. Kwa hiyo,\(\text{C}\) na\(\text{D}\) ni matukio ya kipekee.
    • Hebu tukio\(\text{E} =\) zote nyuso chini ya tano. \(\text{E} = \{1, 2, 3, 4\}\).

    Je\(\text{C}\), na matukio ya\(\text{E}\) kipekee? (Jibu ndiyo au hapana.) Kwa nini au kwa nini?

    Jibu

    Hapana. \(\text{C} = \{3, 5\}\)na\(\text{E} = \{1, 2, 3, 4\}\). \(P(\text{C AND E}) = \dfrac{1}{6}\). Ili kuwa ya kipekee,\(P(\text{C AND E})\) lazima iwe sifuri.

    • Kupata\(P(\text{C|A})\). Hii ni uwezekano wa masharti. Kumbuka kwamba tukio\(\text{C}\) ni {3, 5} na tukio\(\text{A}\) ni {1, 3, 5}. Ili kupata\(P(\text{C|A})\), pata uwezekano wa\(\text{C}\) kutumia nafasi ya sampuli\(\text{A}\). Umepunguza nafasi ya sampuli kutoka nafasi ya awali ya sampuli {1, 2, 3, 4, 5, 6} hadi {1, 3, 5}. Kwa hiyo,\(P(\text{C|A}) = \dfrac{2}{3}\).

    Zoezi\(\PageIndex{5}\)

    Hebu tukio\(\text{A} =\) kujifunza Kihispania. Hebu tukio\(\text{B}\) = kujifunza Kijerumani. Kisha\(\text{A AND B}\) = kujifunza Kihispania na Kijerumani. Tuseme\(P(\text{A}) = 0.4\) na\(P(\text{B}) = 0.2\). \(P(\text{A AND B}) = 0.08\). Je, matukio\(\text{A}\) na\(\text{B}\) kujitegemea? Kidokezo: Lazima uonyeshe MOJAWAPO ya yafuatayo:

    • \(P(\text{A|B}) = P(\text{A})\)
    • \(P(\text{B|A})\)
    • \(P(\text{A AND B}) = P(\text{A})P(\text{B})\)

    Jibu

    \[P(\text{A|B}) = \dfrac{\text{P(A AND B)}}{P(\text{B})} = \dfrac{0.08}{0.2} = 0.4 = P(\text{A})\]

    matukio ni huru kwa sababu\(P(\text{A|B}) = P(\text{A})\).

    Mfano\(\PageIndex{6}\)

    Hebu tukio\(\text{G} =\) kuchukua darasa hisabati. Hebu tukio\(\text{H} =\) kuchukua darasa sayansi. Kisha,\(\text{G AND H} =\) kuchukua darasa math na darasa sayansi. Tuseme\(P(\text{G}) = 0.6\)\(P(\text{H}) = 0.5\),, na\(P(\text{G AND H}) = 0.3\). Je\(\text{G}\), ni\(\text{H}\) huru?

    Ikiwa\(\text{G}\) na\(\text{H}\) ni huru, basi lazima uonyeshe MOJAWAPO ya yafuatayo:

    • \(P(\text{G|H}) = P(\text{G})\)
    • \(P(\text{H|G}) = P(\text{H})\)
    • \(P(\text{G AND H}) = P(\text{G})P(\text{H})\)

    Uchaguzi unayofanya unategemea habari uliyo nayo. Unaweza kuchagua njia yoyote hapa kwa sababu una taarifa muhimu.

    1. a. kuonyesha kwamba\(P(\text{G|H}) = P(\text{G})\).
    2. b. Onyesha\(P(\text{G AND H}) = P(\text{G})P(\text{H})\).

    Suluhisho

    1. \(P(\text{G|H}) = \dfrac{P(\text{G AND H})}{P(\text{H})} = \dfrac{0.3}{0.5} = 0.6 = P(\text{G})\)
    2. \(P(\text{G})P(\text{H}) = (0.6)(0.5) = 0.3 = P(\text{G AND H})\)

    Tangu\(\text{G}\) na\(\text{H}\) ni huru, kujua kwamba mtu anachukua darasa la sayansi haibadili nafasi ya kuwa yeye anachukua darasa la hisabati. Ikiwa matukio hayo mawili hayakuwa huru (yaani, yanategemea) basi kujua kwamba mtu anachukua darasa la sayansi bila kubadilisha nafasi yeye anachukua hesabu. Kwa mazoezi, onyesha kwamba\(P(\text{H|G}) = P(\text{H})\) ili kuonyesha kwamba\(\text{G}\) na\(\text{H}\) ni matukio ya kujitegemea.

    Zoezi\(\PageIndex{6}\)

    Katika mfuko, kuna marumaru sita nyekundu na marumaru manne ya kijani. Marumaru nyekundu ni alama na namba 1, 2, 3, 4, 5, na 6. Marumaru ya kijani ni alama na namba 1, 2, 3, na 4.

    • \(\text{R} =\)marumaru nyekundu
    • \(\text{G} =\)marumaru ya kijani
    • \(\text{O} =\)marumaru isiyo ya kawaida
    • Nafasi ya sampuli ni\(\text{S} = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}\).

    \(\text{S}\)ina matokeo kumi. Ni nini\(P(\text{G AND O})\)?

    Jibu

    Tukio\(\text{G}\) na\(\text{O} = \{G1, G3\}\)

    \(P(\text{G and O}) = \dfrac{2}{10} = 0.2\)

    Mfano\(\PageIndex{7}\)

    Hebu tukio\(\text{C} =\) kuchukua darasa Kiingereza. Hebu tukio\(\text{D} =\) kuchukua darasa hotuba.

    Tuseme\(P(\text{C}) = 0.75\)\(P(\text{D}) = 0.3\),,\(P(\text{C|D}) = 0.75\) na\(P(\text{C AND D}) = 0.225\).

    Thibitisha majibu yako kwa maswali yafuatayo kwa nambari.

    1. Je\(\text{C}\), ni\(\text{D}\) huru?
    2. Je\(\text{C}\), na\(\text{D}\) pande kipekee?
    3. Ni nini\(P(\text{D|C})\)?

    Suluhisho

    1. Ndiyo, kwa sababu\(P(\text{C|D}) = P(\text{C})\).
    2. Hapana, kwa sababu\(P(\text{C AND D})\) si sawa na sifuri.
    3. \(P(\text{D|C}) = \dfrac{P(\text{C AND D})}{P(\text{C})} = \dfrac{0.225}{0.75} = 0.3\)

    Zoezi\(\PageIndex{7}\)

    Mwanafunzi huenda kwenye maktaba. Hebu matukio\(\text{B} =\) mwanafunzi hundi nje kitabu na\(\text{D} =\) mwanafunzi hundi nje DVD. Tuseme kwamba\(P(\text{B}) = 0.40\),\(P(\text{D}) = 0.30\) na\(P(\text{B AND D}) = 0.20\).

    1. Kupata\(P(\text{B|D})\).
    2. Kupata\(P(\text{D|B})\).
    3. Je\(\text{B}\), ni\(\text{D}\) huru?
    4. Je\(\text{B}\), na\(\text{D}\) pande kipekee?

    Jibu

    1. \(P(\text{B|D}) = 0.6667\)
    2. \(P(\text{D|B}) = 0.5\)
    3. Hapana
    4. Hapana

    Mfano\(\PageIndex{8}\)

    Katika sanduku kuna kadi tatu nyekundu na kadi tano za bluu. Kadi nyekundu zimewekwa na namba 1, 2, na 3, na kadi za bluu zimewekwa na namba 1, 2, 3, 4, na 5. Kadi hizo zimefungwa vizuri. Unafikia ndani ya sanduku (huwezi kuona ndani yake) na kuteka kadi moja.

    Hebu

    • \(\text{R =}\)kadi nyekundu hutolewa,
    • \(\text{B} =\)kadi ya bluu hutolewa,
    • \(\text{E} =\)hata-kuhesabiwa kadi ni inayotolewa.

    Nafasi ya sampuli\(S = R1, R2, R3, B1, B2, B3, B4, B5\).

    \(S\)ina matokeo nane.

    • \(P(\text{R}) = \dfrac{3}{8}\). \(P(\text{B}) = \dfrac{5}{8}\). \(P(\text{R AND B}) = 0\). (Huwezi kuteka kadi moja ambayo ni nyekundu na bluu.)
    • \(P(\text{E}) = \dfrac{3}{8}\). (Kuna tatu kadi hata-kuhesabiwa,\(R2, B2\), na\(B4\).)
    • \(P(\text{E|B}) = \dfrac{2}{5}\). (Kuna kadi tano za bluu:\(B1, B2, B3, B4\), na\(B5\). Kati ya kadi za bluu, kuna kadi mbili hata;\(B2\) na\(B4\).)
    • \(P(\text{B|E}) = \dfrac{2}{3}\). (Kuna kadi tatu hata-kuhesabiwa:\(R2, B2\), na\(B4\). Kati ya kadi hata-kuhesabiwa, kwa ni bluu;\(B2\) na\(B4\).)
    • matukio\(\text{R}\) na\(\text{B}\) ni pande kipekee kwa sababu\(P(\text{R AND B}) = 0\).
    • Hebu\(\text{G} =\) kadi na idadi kubwa kuliko 3. \(\text{G} = \{B4, B5\}\). \(P(\text{G}) = \dfrac{2}{8}\). Hebu kadi ya\(\text{H} =\) bluu kuhesabiwa kati ya moja na nne, umoja. \(\text{H} = \{B1, B2, B3, B4\}\). \(P(\text{G|H}) = \frac{1}{4}\). (Kadi tu kwa\(\text{H}\) kuwa ina idadi kubwa kuliko tatu ni B4.) tangu\(\dfrac{2}{8} = \dfrac{1}{4}\),\(P(\text{G}) = P(\text{G|H})\), ambayo ina maana kwamba\(\text{G}\) na\(\text{H}\) ni huru.

    Zoezi\(\PageIndex{8}\)

    Katika uwanja wa mpira wa kikapu,

    • 70% ya mashabiki ni mizizi kwa timu ya nyumbani.
    • 25% ya mashabiki wamevaa bluu.
    • 20% ya mashabiki wamevaa bluu na ni mizizi kwa timu ya mbali.
    • Ya mashabiki wa mizizi kwa timu ya mbali, 67% wamevaa bluu.

    Hebu\(\text{A}\) tuwe tukio ambalo shabiki ni mizizi kwa timu ya mbali.

    Hebu\(\text{B}\) tuwe tukio ambalo shabiki amevaa bluu.

    Je! Matukio ya mizizi kwa timu ya mbali na kuvaa bluu huru? Je, wao ni pande kipekee?

    Jibu

    • \(P(\text{B|A}) = 0.67\)
    • \(P(\text{B}) = 0.25\)

    Hivyo\(P(\text{B})\) si sawa\(P(\text{B|A})\) ambayo ina maana kwamba\(\text{B} and \text{A}\) si huru (amevaa bluu na mizizi kwa timu ya mbali si huru). Pia sio pekee, kwa sababu\(P(\text{B AND A}) = 0.20\), si\(0\).

    Mfano\(\PageIndex{9}\)

    Katika darasa fulani la chuo, 60% ya wanafunzi ni wa kike. Asilimia hamsini ya wanafunzi wote katika darasa wana nywele ndefu. Asilimia arobaini na tano ya wanafunzi ni wa kike na wana nywele ndefu. Kati ya wanafunzi wa kike, 75% wana nywele ndefu. Hebu\(\text{F}\) tuwe tukio ambalo mwanafunzi ni mwanamke. Hebu\(\text{L}\) tuwe tukio ambalo mwanafunzi ana nywele ndefu. Mwanafunzi mmoja huchukuliwa nasibu. Je! Matukio ya kuwa kike na kuwa na nywele ndefu huru?

    • Probabilities zifuatazo zinatolewa katika mfano huu:
    • \(P(\text{F}) = 0.60\);\(P(\text{L}) = 0.50\)
    • \(P(\text{F AND L}) = 0.45\)
    • \(P(\text{L|F}) = 0.75\)

    Uchaguzi unayofanya unategemea habari uliyo nayo. Unaweza kutumia hali ya kwanza au ya mwisho kwenye orodha kwa mfano huu. Hujui\(P(\text{F|L})\) bado, hivyo huwezi kutumia hali ya pili.

    Suluhisho 1

    Angalia kama\(P(\text{F AND L}) = P(\text{F})P(\text{L})\). Sisi ni kutokana na kwamba\(P(\text{F AND L}) = 0.45\), lakini\(P(\text{F})P(\text{L}) = (0.60)(0.50) = 0.30\). Matukio ya kuwa kike na kuwa na nywele ndefu hayajitegemea kwa sababu\(P(\text{F AND L})\) hailingani\(P(\text{F})P(\text{L})\).

    Suluhisho 2

    Angalia kama\(P(\text{L|F})\) ni sawa\(P(\text{L})\). Sisi tumepewa hayo\(P(\text{L|F}) = 0.75\), lakini\(P(\text{L}) = 0.50\) wao si sawa. Matukio ya kuwa kike na kuwa na nywele ndefu sio huru.

    Ufafanuzi wa Matokeo

    Matukio ya kuwa wa kike na kuwa na nywele ndefu hayajitegemea; kujua ya kwamba mwanafunzi ni wa kike hubadilisha uwezekano wa kuwa mwanafunzi ana nywele ndefu.

    Zoezi\(\PageIndex{9}\)

    Mark ni kuamua ni njia gani ya kuchukua kufanya kazi. Uchaguzi wake ni\(\text{I} = \text{the Interstate}\) na\(\text{F} = \text{Fifth Street}\)

    • \(P(\text{I}) = 0.44\)na\(P(\text{F}) = 0.55\)
    • \(P(\text{I AND F}) = 0\)kwa sababu Mark itachukua njia moja tu ya kufanya kazi.

    Je, ni uwezekano wa\(P(\text{I OR F})\) nini?

    Jibu

    Kwa sababu\(P(\text{I AND F}) = 0\),

    \(P(\text{I OR F}) = P(\text{I}) + P(\text{F}) - P(\text{I AND F}) = 0.44 + 0.56 - 0 = 1\)

    Mfano\(\PageIndex{10}\)

    1. Toss sarafu moja ya haki (sarafu ina pande mbili,\(\text{H}\) na\(\text{T}\)). Matokeo ni ________. Hesabu matokeo. Kuna ____ matokeo.
    2. Toss haki moja, sita upande mmoja kufa (kufa ina 1, 2, 3, 4, 5 au 6 dots upande). Matokeo ni ________________. Hesabu matokeo. Kuna ___ matokeo.
    3. Kuzidisha idadi mbili za matokeo. Jibu ni _______.
    4. Kama flip sarafu moja ya haki na kufuata kwa toss ya haki moja, sita upande mmoja kufa, jibu katika tatu ni idadi ya matokeo (ukubwa wa nafasi ya sampuli). Matokeo ni nini? (kidokezo: Mbili ya matokeo ni\(H1\) na\(T6\).)
    5. Tukio\(\text{A} =\) vichwa (\(\text{H}\)) juu ya sarafu ikifuatiwa na idadi hata (2, 4, 6) juu ya kufa.
      \(\text{A}\)= {_________________}. Kupata\(P(\text{A})\).
    6. Tukio\(\text{B} =\) vichwa juu ya sarafu ikifuatiwa na tatu juu ya kufa. \(\text{B} =\){________}. Kupata\(P(\text{B})\).
    7. Je\(\text{A}\), na\(\text{B}\) pande kipekee? (Kidokezo: Ni nini\(P(\text{A AND B})\)? Ikiwa\(P(\text{A AND B}) = 0\), basi\(\text{A}\) na\(\text{B}\) ni ya kipekee.)
    8. Je\(\text{A}\), ni\(\text{B}\) huru? (Kidokezo: Je\(P(\text{A AND B}) = P(\text{A})P(\text{B})\)? Ikiwa\(P(\text{A AND B})\ = P(\text{A})P(\text{B})\), basi\(\text{A}\) na\(\text{B}\) ni huru. Ikiwa sio, basi wanategemea).

    Suluhisho

    1. \(\text{H}\)na\(\text{T}\); 2
    2. 1, 2, 3, 4, 5, 6;
    3. 2 (6) = 12
    4. \(T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6\)
    5. \(\text{A} = \{H2, H4, H6\}\);\(P(\text{A}) = \dfrac{3}{12}\)
    6. \(\text{B} = \{H3\}\);\(P(\text{B}) = \dfrac{1}{12}\)
    7. Ndiyo, kwa sababu\(P(\text{A AND B}) = 0\)
    8. \(P(\text{A AND B}) = 0\). \(P(\text{A})P(\text{B}) = \left(\dfrac{3}{12}\right)\left(\dfrac{1}{12}\right)\). \(P(\text{A AND B})\)haina sawa\(P(\text{A})P(\text{B})\), hivyo\(\text{A}\) na\(\text{B}\) ni tegemezi.

    Zoezi\(\PageIndex{10}\)

    Sanduku lina mipira miwili, nyeupe moja na nyekundu moja. Sisi kuchagua mpira mmoja, kuiweka nyuma katika sanduku, na kuchagua mpira wa pili (sampuli na uingizwaji). Hebu\(\text{T}\) tuwe tukio la kupata mpira mweupe mara mbili, tukio\(\text{F}\) la kuokota mpira mweupe kwanza, tukio\(\text{S}\) la kuokota mpira mweupe katika kuchora pili.

    1. kukokotoa\(P(\text{T})\).
    2. kukokotoa\(P(\text{T|F})\).
    3. Je\(\text{T}\), na\(\text{F}\) kujitegemea?.
    4. Je\(\text{F}\), na\(\text{S}\) pande kipekee?
    5. Je\(\text{F}\), ni\(\text{S}\) huru?

    Jibu

    1. \(P(\text{T}) = \dfrac{1}{4}\)
    2. \(P(\text{T|F}) = \dfrac{1}{2}\)
    3. Hapana
    4. Hapana
    5. Ndio

    Marejeo

    1. Lopez, Shane, Preety Sidhu. “Walimu wa Marekani Upendo Maisha yao, lakini Mapambano katika sehemu za kazi.” Gallup Ustawi, 2013. http://www.gallup.com/poll/161516/te...workplace.aspx (ilifikia Mei 2, 2013).
    2. Takwimu kutoka Gallup. Inapatikana mtandaoni kwenye www.gallup.com/ (imefikia Mei 2, 2013).

    Tathmini

    Matukio mawili\(\text{A}\) na\(\text{B}\) ni huru kama ujuzi kwamba moja ilitokea hauathiri nafasi nyingine hutokea. Ikiwa matukio mawili hayategemea, basi tunasema kuwa wanategemea.

    Katika sampuli na uingizwaji, kila mwanachama wa idadi ya watu hubadilishwa baada ya kuchukuliwa, hivyo mwanachama huyo ana uwezekano wa kuchaguliwa zaidi ya mara moja, na matukio yanachukuliwa kuwa huru. Katika sampuli bila uingizwaji, kila mwanachama wa idadi ya watu anaweza kuchaguliwa mara moja tu, na matukio yanachukuliwa kuwa si huru. Wakati matukio hayashiriki matokeo, wao ni wa kipekee wa kila mmoja.

    Mapitio ya Mfumo

    • Ikiwa\(\text{A}\) na\(\text{B}\) ni huru,\(P(\text{A AND B}) = P(\text{A})P(\text{B}), P(\text{A|B}) = P(\text{A})\) na\(P(\text{B|A}) = P(\text{B})\).
    • Kama\(\text{A}\) na\(\text{B}\) ni pande kipekee,\(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) and P(\text{A AND B}) = 0\).

    Zoezi\(\PageIndex{11}\)

    \(\text{E}\)na\(\text{F}\) ni matukio pande kipekee. \(P(\text{E}) = 0.4\);\(P(\text{F}) = 0.5\). Kupata\(P(\text{E∣F})\).

    Zoezi\(\PageIndex{12}\)

    \(\text{J}\)na\(\text{K}\) ni matukio ya kujitegemea. \(P(\text{J|K}) = 0.3\). Kupata\(P(\text{J})\).

    Jibu

    \(P(\text{J}) = 0.3\)

    Zoezi\(\PageIndex{13}\)

    \(\text{U}\)na\(\text{V}\) ni matukio pande kipekee. \(P(\text{U}) = 0.26\);\(P(\text{V}) = 0.37\). Kupata:

    1. \(P(\text{U AND V}) =\)
    2. \(P(\text{U|V}) =\)
    3. \(P(\text{U OR V}) =\)

    Zoezi\(\PageIndex{14}\)

    \(\text{Q}\)na\(\text{R}\) ni matukio ya kujitegemea. \(P(\text{Q}) = 0.4\)na\(P(\text{Q AND R}) = 0.1\). Kupata\(P(\text{R})\).

    Jibu

    \(P(\text{Q AND R}) = P(\text{Q})P(\text{R})\)

    \(0.1 = (0.4)P(\text{R})\)

    \(P(\text{R}) = 0.25\)

    Kuleta Pamoja

    Zoezi\(\PageIndex{16}\)

    Mwaka uliopita, uzito wa wanachama wa San Francisco 49ers na Dallas Cowboys zilichapishwa katika San Jose Mercury News. Data sahihi imeandaliwa katika Jedwali.

    Shati # ≤ 210 211—250 251—290 290≤
    1—33 21 5 0 0
    34—66 6 18 7 4
    66—99 6 12 22 5

    Kwa yafuatayo, tuseme kwamba nasibu kuchagua mchezaji mmoja kutoka 49ers au Cowboys.

    Ikiwa kuwa na namba ya shati kutoka moja hadi 33 na uzito wa paundi 210 zilikuwa matukio ya kujitegemea, basi ni nini kinachopaswa kuwa kweli kuhusu\(P(\text{Shirt} \#1–33|\leq 210 \text{ pounds})\)?

    Zoezi\(\PageIndex{17}\)

    Uwezekano kwamba kiume anaendelea aina fulani ya kansa katika maisha yake ni 0.4567. Uwezekano wa kuwa mwanamume ana angalau matokeo moja ya mtihani chanya ya uongo (maana ya mtihani unarudi kwa saratani wakati mtu hana hiyo) ni 0.51. Baadhi ya maswali yafuatayo hawana taarifa za kutosha kwa wewe kujibu. Andika “habari zisizo za kutosha” kwa majibu hayo. Hebu\(\text{C} =\) mtu aendelee kansa katika maisha yake na\(\text{P} =\) mtu ana angalau chanya moja ya uongo.

    1. \(P(\text{C}) =\)______
    2. \(P(\text{P|C}) =\)______
    3. \(P(\text{P|C'}) =\)______
    4. Ikiwa mtihani unakuja chanya, kulingana na maadili ya namba, unaweza kudhani kwamba mtu ana kansa? Thibitisha numerically na kueleza kwa nini au kwa nini.

    Jibu

    1. \(P(\text{C}) = 0.4567\)
    2. habari haitoshi
    3. habari haitoshi
    4. Hapana, kwa sababu zaidi ya nusu (0.51) ya wanaume wana angalau maandishi mazuri ya uongo

    Zoezi\(\PageIndex{18}\)

    Kutokana\(\text{G}\) na matukio na\(\text{H}: P(\text{G}) = 0.43\)\(P(\text{H}) = 0.26\);\(P(\text{H AND G}) = 0.14\)

    1. Kupata\(P(\text{H OR G})\).
    2. Kupata uwezekano wa inayosaidia ya tukio (\(\text{H AND G}\)).
    3. Kupata uwezekano wa inayosaidia ya tukio (\(\text{H OR G}\)).

    Zoezi\(\PageIndex{19}\)

    Kutokana\(\text{J}\) na matukio na\(\text{K}: P(\text{J}) = 0.18\)\(P(\text{K}) = 0.37\);\(P(\text{J OR K}) = 0.45\)

    1. Kupata\(P(\text{J AND K})\).
    2. Kupata uwezekano wa inayosaidia ya tukio (\(\text{J AND K}\)).
    3. Kupata uwezekano wa inayosaidia ya tukio (\(\text{J AND K}\)).

    Jibu

    1. \(P(\text{J OR K}) = P(\text{J}) + P(\text{K}) − P(\text{J AND K}); 0.45 = 0.18 + 0.37 - P(\text{J AND K})\); kutatua kupata\(P(\text{J AND K}) = 0.10\)
    2. \(P(\text{NOT (J AND K)}) = 1 - P(\text{J AND K}) = 1 - 0.10 = 0.90\)
    3. \(P(\text{NOT (J OR K)}) = 1 - P(\text{J OR K}) = 1 - 0.45 = 0.55\)

    faharasa

    Matukio tegemezi
    Ikiwa matukio mawili hayategemea, basi tunasema kuwa wanategemea.
    Sampuli na uingizwaji
    Ikiwa kila mwanachama wa idadi ya watu hubadilishwa baada ya kuchukuliwa, basi mwanachama huyo ana uwezekano wa kuchaguliwa zaidi ya mara moja.
    Sampuli bila Replacement
    Wakati sampuli imefanywa bila uingizwaji, kila mwanachama wa idadi ya watu anaweza kuchaguliwa mara moja tu.
    Uwezekano wa masharti ya Tukio moja Kutokana na Tukio Lingine
    P (A | B) ni uwezekano kwamba tukio A litatokea kutokana na kwamba tukio B tayari limetokea.
    AU ya Matukio mawili
    Matokeo ni katika tukio A AU B ikiwa matokeo ni katika A, iko katika B, au iko katika A na B.