Question

A bag contains 6 balls Two balls are drawn from it at random and both are found to be black The probability that the bag contains at least 5 black balls is

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{5}{7}$
• D. $\frac{5}{6}$

Hint:

When calculating probabilities with combinations, ensure to correctly identify the favorable outcomes and divide them by the total possible outcomes.

Answer 1

The Correct Option is C

We are given a bag containing 6 balls, and two balls are drawn at random. The problem asks us to find the probability that the bag contains at least 5 black balls, given the probability that both balls drawn are black. 
Let's define the events: 
- The total number of ways to choose 2 balls from 6 is given by \( \binom{6}{2} \). 
- The probability that both balls drawn are black depends on the number of black balls in the bag.
We need to calculate the number of ways to choose 2 black balls from the possible number of black balls. We are considering cases where there are 5 or 6 black balls in the bag. 
The probability that both balls drawn are black is calculated for each of these cases. \[ \frac{\binom{5}{2} + \binom{6}{2}}{\binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} + \binom{6}{2} + \binom{8}{2}} = \frac{10 + 15}{1 + 3 + 6 + 10 + 15} \] Now, simplifying the fraction: \[ = \frac{25}{35} = \frac{5}{7} \] Thus, the probability that the bag contains at least 5 black balls is \( \frac{5}{7} \). This approach involves calculating the total number of favorable outcomes (selecting 2 black balls) and dividing it by the total possible outcomes (selecting any 2 balls). We can also apply the principle of conditional probability to further understand how this result relates to the actual composition of the bag.