A basketball game is played between team Blue and Red. There are a total of 9 players in each them and 5 will play in the game. Ankit is in team blue and Vaibhav is in team Red. What is the probability that at least one of ankit or vaibhave is in playing five?

To solve this problem, we first need to calculate the probability of Ankit and Vaibhav each being in the playing five of their respective teams, and then use these probabilities to find the probability of at least one of them being in the playing five.

Step 1: Calculate the probability of Ankit being in the playing five of team Blue.

The total number of ways to select 5 players out of 9 is given by the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of players, r is the number of players to be selected, and '!' denotes factorial.

So, the total number of ways to select 5 players out of 9 is C(9, 5) = 9! / [5!(9-5)!] = 126.

Since Ankit is one player, the number of ways to select the remaining 4 players out of the remaining 8 is C(8, 4) = 8! / [4!(8-4)!] = 70.

So, the probability of Ankit being in the playing five is 70/126 = 0.556.

Step 2: Calculate the probability of Vaibhav being in the playing five of team Red.

The calculation is the same as for Ankit, so the probability is also 0.556.

Step 3: Calculate the probability of at least one of Ankit or Vaibhav being in the playing five.

The probability of at least one of two events happening is given by the formula P(A or B) = P(A) + P(B) - P(A and B).

In this case, the events are not mutually exclusive, so we need to subtract the probability of both Ankit and Vaibhav being in the playing five.

The probability of both Ankit and Vaibhav being in the playing five is the product of their individual probabilities, which is 0.556 * 0.556 = 0.309.

So, the probability of at least one of Ankit or Vaibhav being in the playing five is 0.556 + 0.556 - 0.309 = 0.803, or approximately 80.3%.