Q77 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

Bag $I$ contains $3$ red and $4$ black balls. Bag II contains $5$ red and $6$ black balls. If one ball is drawn at random from one of the bags, and it is found to be red, then the probability that it was drawn from Bag $II$ , is

A
$\frac{33}{68}$
Correct answer
B
$\frac{35}{68}$
C
$\frac{37}{68}$
D
$\frac{41}{68}$

Solution

Let, $U_{1} & U_{2}$ be the events of the selection of bag $I$ and bag $II$ , respectively.

$\Rightarrow P (U_{1}) = P (U_{2}) = \frac{1}{2}$

Now, $P (\frac{U_{2}}{R}) = \frac{P (U_{2} \cap R)}{P (R)}$

$\Rightarrow P (\frac{U_{2}}{R}) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{35}{68}$

Hence, the correct answer is option: 'B'.

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Bag I contains 3 red and 4 black balls. Bag II contains 5 red and 6 black balls. If one ball is drawn at random from one of the bags, and it is found to be red, then the probability that it was drawn from Bag II, is

Solution

Relevant Questions

Bag $I$ contains $3$ red and $4$ black balls. Bag II contains $5$ red and $6$ black balls. If one ball is drawn at random from one of the bags, and it is found to be red, then the probability that it was drawn from Bag $II$ , is