Question

If a fair coin is tossed 6 times, what is the probability to get exactly 2 heads? at most 2 heads?

Answer 1

The probabilities are:

exactly 2 heads: `P(A)=15/64`

at most 2 heads: `P(B)=11/32`

Explanation:

In this task you can use the rule called Bernoulli's Scheme.

In this scheme you repeat an experiment which can end with one of 2 results (usually called a success and a failure) and want to calculate the probability of getting exactly `k` "success" results.

The probability can be calculated as:

`P(S_k)=((n),(k))p^k(1-p)^(n-k)`, where:

`n` is a total number of experiments
`k` is an expected number of successes
`p` is the probability of a success

The first task can be simply solved by calculating the probability of 2 successes:

A "success" is "tossing heads in a single toss"
A "failure" is "tossing tails in a single toss"

The probability of a success is `1/2`

The number of all experiments is `n=6`

The number of successes is `k=2`

`P(S_2)=((6),(2))(1/2)^2(1/2)^(4)=((6),(2))*(1/2)^6=(6!)/(4!*2!)(1/64)=15/64`

In the second part you have to calculate `P(S_2)+P(S_1)+P(S_0)`

`P(S_0)+P(S_1)+P(S_2)=`

`((6),(0))(1/2)^6+((6),(1))(1/2)^6+((6),(2))(1/2)^6=`

`1*1/64+6*1/64+15*1/64=22/64=11/32`