(a) A ⊂ B, P(A) ≠ 0 are two events, A and B, then P(A/B) equals
(a) P(A)/P(B) (b) P(B)/P(A) (c) 1/P(A) (d) 1/P(B) (Score : 1)
(b) There are two identical bags. Bag I has 3 red and 4 black balls. Bag II has 5 red and 4 black balls. A ball is drawn at random from one of the bags.
(

Question

(a) A ⊂ B, P(A) ≠ 0 are two events, A and B, then P(A/B) equals
(a) P(A)/P(B) (b) P(B)/P(A) (c) 1/P(A) (d) 1/P(B) (Score : 1)
(b) There are two identical bags. Bag I has 3 red and 4 black balls. Bag II has 5 red and 4 black balls. A ball is drawn at random from one of the bags.
(i) What is the probability that the ball drawn is red ? (Scores : 2)
(ii) If the ball drawn is red, what is the probability that it is from Bag I ? (Scores : 2)

**OR**

X is a random variable whose probability distribution is given below :
X : 0, 1, 2, 3, 4
P(X) : 1/16, 2/16, K, 5/16, 1/16
(i) Find the value of K. (Score : 1)
(ii) Find the Mean and Variance of X. (Scores : 4)

Accepted Answer

>

> [!TLDR]
> Main uses conditional probability and Bayes' theorem for the two-bag problem; OR uses the discrete probability distribution's normalisation condition and the formulas for mean and variance.
>
> **(Main) (a) $A\subset B$, $P(A)
> eq0$. Find $P(A/B)$.**
>
> $P(A/B) = \dfrac{P(A\cap B)}{P(B)}$. Since $A\subset B$, $A\cap B = A$, so $P(A\cap B)=P(A)$.
>
> $$P(A/B) = \frac{P(A)}{P(B)}$$ — option (a).
>
> **(Main) (b) Bag I: 3 red, 4 black. Bag II: 5 red, 4 black. A bag is chosen at random, then a ball is drawn.**
>
> *(i) $P(	ext{red})$:*
> …

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