Question

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$, where gcd(m, n) = 1, then m + n is equal to :

• A. 4
• B. 14
• C. 13
• D. 11

Answer 1

Answer: (B)

14

Answer 2

To solve this problem, we need to calculate the conditional probability P(first ball is black | second ball is black).

Let's define the events:

• Let $B_1$ be the event that the first selected ball is black.
• Let $B_2$ be the event that the second selected ball is black.
• Let $W_1$ be the event that the first selected ball is white.

We are asked to find $P(B_1 | B_2)$. Using the formula for conditional probability, we have: $P(B_1 | B_2) = \frac{P(B_1 \cap B_2)}{P(B_2)}$

The bag contains 4 white balls and 6 black balls, for a total of 10 balls.

Step 1: Calculate $P(B_1 \cap B_2)$

This is the probability that the first ball is black AND the second ball is black.

• Probability of the first ball being black, $P(B_1) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{6}{10}$.
• After selecting one black ball without replacement, there are 5 black balls left and a total of 9 balls.
• Probability of the second ball being black given the first was black, $P(B_2 | B_1) = \frac{5}{9}$.
• Therefore, $P(B_1 \cap B_2) = P(B_1) \times P(B_2 | B_1) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.

Step 2: Calculate $P(B_2)$

The second ball can be black in two mutually exclusive ways:

1. The first ball was black AND the second ball is black ($B_1 \cap B_2$).
2. The first ball was white AND the second ball is black ($W_1 \cap B_2$).

We already calculated $P(B_1 \cap B_2) = \frac{1}{3}$.

Now, let's calculate $P(W_1 \cap B_2)$:

• Probability of the first ball being white, $P(W_1) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10}$.
• After selecting one white ball without replacement, there are still 6 black balls left and a total of 9 balls.
• Probability of the second ball being black given the first was white, $P(B_2 | W_1) = \frac{6}{9}$.
• Therefore, $P(W_1 \cap B_2) = P(W_1) \times P(B_2 | W_1) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$.

Now, sum these probabilities to get $P(B_2)$: $P(B_2) = P(B_1 \cap B_2) + P(W_1 \cap B_2) = \frac{1}{3} + \frac{4}{15}$ To add these fractions, find a common denominator, which is 15: $\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$ So, $P(B_2) = \frac{5}{15} + \frac{4}{15} = \frac{9}{15} = \frac{3}{5}$.

Step 3: Calculate $P(B_1 | B_2)$

Now we can use the conditional probability formula: $P(B_1 | B_2) = \frac{P(B_1 \cap B_2)}{P(B_2)} = \frac{\frac{1}{3}}{\frac{3}{5}}$ $P(B_1 | B_2) = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9}$.

The probability is given as $\frac{m}{n}$, where $\text{gcd}(m, n) = 1$. Here, $m = 5$ and $n = 9$. $\text{gcd}(5, 9) = 1$, so the condition is satisfied.

Finally, we need to find $m + n$: $m + n = 5 + 9 = 14$.