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Q.State Bayes' theorem on probability. Use this theorem to solve any one of the following (1+5=6):

(a) Bag II contains 3 red and 4 black balls and Bag IIII contains 4 red and 5 black balls. One ball is transferred from Bag II to Bag IIII and then one ball is drawn from Bag IIII. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. OR
(b) A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2025Subjective· 6mImportance★★★★★
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State Bayes' theorem, set up the prior and likelihoods, then compute the posterior.

Bayes' theorem. If E1,E2,…,EnE_1,E_2,\dots,E_n are mutually exclusive and exhaustive events with P(Ei)>0P(E_i)>0, and AA is any event with P(A)>0P(A)>0, then

P(Ei∣A)=P(Ei) P(A∣Ei)∑j=1nP(Ej) P(A∣Ej).P(E_i\mid A)=\frac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}.

(a) Bag I has 3 red + 4 black (7 balls); Bag II has 4 red + 5 black (9 balls). One ball is transferred I→II, then a ball is drawn from Bag II and found red (AA).

Let E1E_1 = transferred ball is red, E2E_2 = transferred ball is black. Priors:

P(E1)=37,P(E2)=47.P(E_1)=\frac37,\qquad P(E_2)=\frac47.

After transfer, Bag II holds 10 balls:

  • If red transferred: 5 red, 5 black ⇒P(A∣E1)=510=12.\Rightarrow P(A\mid E_1)=\dfrac{5}{10}=\dfrac12.
  • If black transferred: 4 red, 6 black ⇒P(A∣E2)=410=25.\Rightarrow P(A\mid E_2)=\dfrac{4}{10}=\dfrac25.

Apply Bayes' theorem for the black-transfer case:

P(E2∣A)=P(E2)P(A∣E2)P(E1)P(A∣E1)+P(E2)P(A∣E2)=47⋅2537⋅12+47⋅25.P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}=\frac{\frac47\cdot\frac25}{\frac37\cdot\frac12+\frac47\cdot\frac25}.

Numerator =835=\dfrac{8}{35}. Denominator =314+835=1570+1670=3170=\dfrac{3}{14}+\dfrac{8}{35}=\dfrac{15}{70}+\dfrac{16}{70}=\dfrac{31}{70}. Since 835=1670\dfrac{8}{35}=\dfrac{16}{70},

P(E2∣A)=16/7031/70=1631.P(E_2\mid A)=\frac{16/70}{31/70}=\frac{16}{31}.

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