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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.

User Moin
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1 Answer

3 votes

Answer:


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

User Ijt
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