A fair coin is tossed 6 times. What is the probability that exactly 2

We can treat this as an ARRANGEMENT or a SELECTION problem. Let us see how:

Approach 1:

We need to have H H T T T T (2 Heads and 4 Tails)

This can be treated as an ARRANGEMENT Problem:

Number of ways of arranging 2 H and 4 T = \(\frac{6!}{(2! * 4!)}\) = \(\frac{(6 * 5 * 4!)}{(2 * 4!)}\) = 15

Total number of ways in which 6 coins can be tossed (where each coin has 2 options - either Head or Tail) = \(2^6 = 64\)

Required probability = \(\frac{15}{64}\)

Answer D


Approach 2:

We need 2 Heads (and hence, 4 Tails)

We can treat this as a SELECTION problem where we need to select 2 Heads from 6 coins

Number of ways of selecting = \(6C2\) = \(\frac{6!}{(2! * 4!)}\) = 15

Total number of ways in which 6 coins can be tossed (where each coin has 2 options - either Head or Tail) = \(2^6 = 64\)

Required probability = \(\frac{15}{64}\)

Answer D

6C2 *(1/2)^2*(1/2)^4 = 15*(1/64) = 15/64

Given that A fair coin is tossed 6 times and we need to find What is the probability that exactly 2 heads will show.

Coin is tossed 6 times => Total number of cases = \(2^6\) = 64

Cases in which we will get 2 Heads

Cases in which we get 2H can be found by putting 2 heads in any of the 6 slots _ _ _ _ _ _ _
This can be done in 6C2 ways = \(\frac{6!}{2!*(6-2)!}\) = \(\frac{6*5*4!}{2!*4!}\) = 15

=> P(6H) = \(\frac{15}{64}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems