Question

A coin is tossed 7 times. what is the probability that the number of heads obtained will be between 22 and 77 inclusive? express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

`P(2\leq X\leq 7)= 0.9375`

`P(2<X<7)= 0.7656`

Explanation:

Experimeto is to throw a coin 7 times and count the number of faces obtained.

So, getting a success means getting a face.

The probability p of obtaining success is:

`p = 0.5`

The probability of obtaining a failure is:

`q = 0.5`

Then we have all the necessary conditions to have a binomial experiment. Therefore we use the following formula:

`P(X) =(n!)/(x!(n-x)!)p^xq^(n-x)`

Where x represents the number of successes. In this case we want the probability that:

`2\leq X\leq 7\\\\P(2\leq X\leq 7) = 1- P(x\leq 1)\\\\P(x\leq 1) = P(0) + P(1)\\\\\\= (7!)/(0!(7-0)!)(0.5)^0(0.5)^(7-0) + (7!)/(1!(7-1)!)(0.5)^1(0.5)^(7-1)\\\\\\P(x\leq 1) = (1)/(128) + (7)/(128) = (8)/(128) = (1)/(16)\\\\P(2\leq X\leq 7) = 1- (1)/(16)\\\\P(2\leq X\leq 7) = (15)/(16) = 0.9375`

Now assuming that the desired probability is:

`P(2<X<7)`

So:

`P(2<X<7) = P(3) + P(4) + P(5) + P(6)`

=
`\sum^(6)_(x=3)\ (7!)/(x!(7-x)!)(0.5)^x(0.5)^(7-x)`

`= (49)/(64) = 0.7656`

Finally

`P(2<X<7)= 0.7656`