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:
Show Hint
- 14
- 4
- 11
- 13
The Correct Option is A
Approach Solution - 1
To solve the problem, we need to find the probability \(P(A|B)\), where event \(A\) is that the first ball is black, and event \(B\) is that the second ball is black. The probability formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Step 1: Find \(P(A \cap B)\).
The probability that the first ball is black and the second ball is also black can be calculated by considering the following:
- Number of ways to select the first black ball: 6 (out of 10 total balls).
- After selecting the first black ball, 5 black balls remain (and 9 total balls).
- Number of ways to select the second black ball: 5 (out of 9 total balls).
Therefore:
\[ P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \]
Step 2: Find \(P(B)\).
\(P(B)\) is the probability that the second ball is black regardless of the color of the first ball. Consider the two scenarios:
- First ball is black (probability): \[ \frac{6}{10} \]
- If first ball is black, probability that second is black: \[ \frac{5}{9} \]
- First ball is white (probability): \[ \frac{4}{10} \]
- If first ball is white, probability that second is black: \[ \frac{6}{9} \]
Thus:
\[ P(B) = \left(\frac{6}{10} \times \frac{5}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac{3}{5} \]
Step 3: Calculate \(P(A|B)\).
Using the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \]
With gcd(5, 9) = 1, the fraction \(\frac{m}{n}\) is in its simplest form with \(m = 5\) and \(n = 9\). Hence, \(m + n\) equals \(14\).
Approach Solution -2
Let
$B_1$ : event that the first ball is black
$B_2$ : event that the second ball is black
We need to find $\displaystyle P(B_1 \mid B_2) = \dfrac{P(B_1 \cap B_2)}{P(B_2)}$
Step 1: Find $P(B_1 \cap B_2)$
The probability that both balls are black is \[ P(B_1 \cap B_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \]
Step 2: Find $P(B_2)$
The second ball can be black in two ways:
- First black, second black: $\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90}$
- First white, second black: $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}$
Hence, \[ P(B_2) = \frac{30 + 24}{90} = \frac{54}{90} = \frac{3}{5} \]
Step 3: Find $P(B_1 \mid B_2)$ \[ P(B_1 \mid B_2) = \frac{P(B_1 \cap B_2)}{P(B_2)} = \frac{\tfrac{1}{3}}{\tfrac{3}{5}} = \frac{5}{9} \]
Thus, $\displaystyle \frac{m}{n} = \frac{5}{9}$ and since $\gcd(5,9)=1$, \[ m+n = 5+9 = 14 \]
Final Answer:
$\boxed{m+n = 14}$
Correct Option: 1
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