Question

What are the probabilities that...

1) You toss a fair coin 6 times and you get exactly 4 heads?

2) You toss a fair coin 6 times and you get at least two heads?

Answer 1

Solution:

Let X be random variable which represents the number of heads obtained when tossing a fair coin 6 times.

Probability of obtaining a head on tossing a fair coin is 1/2 = 0.5

Let us consider "getting head" as success. Hence, we have only two mutually exclusive outcomes.

Probability of success (p) = 0.5

Number of trials (n) = 6

Since, number of trials are finite, probability of success remains constant in each of the trials, the outcomes are independent and we have only two mutually exclusive outcomes for each of the trials, therefore we can consider that X follows binomial distribution.

According to binomial probability law, probability of occurrence of exactly x successes in n trials is given by,

Where, p is probability of success.

1) We have to obtain P(X = 4).

We have, n = 6 and p = 0.5

Hence, the probability that I get exactly 4 heads is 0.2344.

2) We have to obtain P(X = at least 2).

We have, n = 6 and p = 0.5

P(X = at least 2) = P(X ≥ 2)

P(X ≥ 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)]

P(X = at least 2) = 0.8906

Hence, the probability that I get at least 2 heads is 0.8906.

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