Five coins are tossed once. What is the probability of getting at most four tails ?

Concept:

If A and B are independent, then P(A and B) = P(A).P(B)

Explanation:

Probability of getting atmost four tails in five tosses = 1 - Probability of getting all five tails

⇒ Probability of getting atmost four tails in five tosses = 1 - P(T,T,T,T,T)

In one toss, P(H) = \(1 \over 2\) and P(T) = \(1 \over 2\)

Since all five tosses are independent,

⇒ P(T,T,T,T,T) = \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) = \(1 \over 32\)

⇒ Probability of getting atmost four tails in five tosses = 1 - ​​\(1 \over 32\) = ​​​\(31 \over 32\)​

∴ The correct option is (1).