A bag consist of 4 red, 5 blue and 6 green coloured socks. If 3 socks are selected, what is the probability that exact one of them is blue?Question 1Answera.15/23b.45/91c.5/9d.13/56

The total number of socks is 4 (red) + 5 (blue) + 6 (green) = 15 socks.

We want to find the probability that exactly one of the three socks drawn is blue. This can happen in three ways:

1. The first sock is blue, and the other two are not blue.
2. The second sock is blue, and the other two are not blue.
3. The third sock is blue, and the other two are not blue.

Let's calculate the probability for each case:

1. The probability that the first sock is blue is 5/15. The probability that the second sock is not blue is 10/14 (since we have 10 non-blue socks left and 14 socks in total). The probability that the third sock is not blue is 9/13 (since we have 9 non-blue socks left and 13 socks in total). So, the total probability for this case is (5/15) * (10/14) * (9/13).

2. The probability that the first sock is not blue is 10/15. The probability that the second sock is blue is 5/14 (since we have 5 blue socks left and 14 socks in total). The probability that the third sock is not blue is 9/13 (since we have 9 non-blue socks left and 13 socks in total). So, the total probability for this case is (10/15) * (5/14) * (9/13).

3. The probability that the first sock is not blue is 10/15. The probability that the second sock is not blue is 9/14 (since we have 9 non-blue socks left and 14 socks in total). The probability that the third sock is blue is 5/13 (since we have 5 blue socks left and 13 socks in total). So, the total probability for this case is (10/15) * (9/14) * (5/13).

Since these are mutually exclusive events, we can add the probabilities together to get the total probability.

So, the total probability that exactly one of the three socks drawn is blue is (5/15) * (10/14) * (9/13) + (10/15) * (5/14) * (9/13) + (10/15) * (9/14) * (5/13) = 45/91.

So, the correct answer is b. 45/91.