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

Answer:

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

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

Step-by-step 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:

$P(X) =\frac{n!}{x!(n-x)!}p^xq^{n-x}$

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

$2\leq X\leq 7\\\\P(2\leq X\leq 7) = 1- P(x\leq 1)\\\\P(x\leq 1) = P(0) + P(1)\\\\\\= \frac{7!}{0!(7-0)!}(0.5)^0(0.5)^{7-0} + \frac{7!}{1!(7-1)!}(0.5)^1(0.5)^{7-1}\\\\\\P(x\leq 1) = \frac{1}{128} + \frac{7}{128} = \frac{8}{128} = \frac{1}{16}\\\\P(2\leq X\leq 7) = 1- \frac{1}{16}\\\\P(2\leq X\leq 7) = \frac{15}{16} = 0.9375$

Now assuming that the desired probability is:

$P(2<X<7)$

So:

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

=  $\sum^{6}_{x=3}\ \frac{7!}{x!(7-x)!}(0.5)^x(0.5)^{7-x}$

$= \frac{49}{64} = 0.7656$

Finally

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