Getting at-least 4 Heads out of 6 = exactly 4H + exactly 5H + all 6H

exactly 4H = \(^6C_4\)
exactly 5H = \(^6C_5\)
All 6H = \(^6C_6\)

Therefore,
Total ways of getting at-least 4 Heads out of 6 = 15 + 6 + 1 = 22
Total possible outcomes = \(2^6\) = 64

Required Probability = \(\frac{22}{64}\) = \(\frac{11}{32}\)

Hence, option D

Given that A coin is tossed six times and we need to find What is the probability of getting at least four heads on the tosses?

At least 4 heads means that we can get 4 or more heads

=> P(At least 4 heads) = P(4H) + P(5H) + P(6H)

P(4H)

Total number of cases = \(2^6\) = 64
Case in which we get 4 heads can be found by arranging 4 heads in _ _ _ _ _ _ 6 spaces.
=> 6C4 ways => \(\frac{6!}{4!*(6-4)!}\) = \(\frac{6!}{4!*2!}\) = \(\frac{6*5*4!}{4!*2!}\) = 15 ways

=> P(4H) = \(\frac{15}{64}\)

P(5H)

Case in which we get 5 heads can be found by arranging 5 heads in _ _ _ _ _ _ 6 spaces.
=> 6C5 ways => \(\frac{6!}{5!*(6-5)!}\) = \(\frac{6!}{5!*1!}\) = \(\frac{6*5!}{5!*1!}\) = 6 ways

=> P(5H) = \(\frac{6}{64}\)

P(6H)

There is only one case where we can get all 6 values as heads. (HHHHHH)

=> P(6H) = \(\frac{1}{64}\)

=> P(At least 4H) = P(4H) + P(5H) + P(6H) = \(\frac{15}{64}\) + \(\frac{6}{64}\) + \(\frac{1}{64}\) = \(\frac{22}{64}\) = \(\frac{11}{32}\)

So, Answer will be D
Hope it helps!

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

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