First slide

Theorems of probability

Question

In a box, there are 20 cards, out of which 10 are labelled as A and the remaining 10 are labelled as B. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is:

Moderate

A

$\frac{15}{16}$
B

$\frac{11}{16}$
C

$\frac{13}{16}$
D

$\frac{9}{16}$

Solution

$The possibilities are (A A), (A, B, A), (A, B, B, A)$
$The probability of getting A A is \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

$\begin{array}{l} The possibilities to the second choice (A, B, B, A) are (A, B, B, A) or \\ (B, A, B, A), (B, B, A, A), (A, B, B, A) \end{array}$
$\begin{array}{l} The probability of getting (A, B, B, A) is 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{16} \\ Therefore, the required probability will be \frac{1}{4} + \frac{1}{4} + \frac{3}{16} = \frac{11}{16} \end{array}$
$\begin{array}{l} Hint: The possibilities that the second A-card appears before the third B-card are \\ (A A), (A, B, A), (A, B, B, A) \end{array}$

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