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Based on a Comcast​ survey, there is a 0.8 probability that a randomly selected adult will watch​ prime-time TV​ live, instead of​ online, on​ DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch​ prime-time live.

A. 0.00430
B. 0.00467
C. 0.000358
D. 0.0512

User Jotaen
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Answer:

B. 0.00467

Explanation:

This is a binomial probability problem. The probability of fewer than three adults watching prime-time TV live is the sum of the probabilities of 0, 1, and 2 adults watching prime-time TV live.

Let X be the number of adults watching prime-time TV live. The probability mass function of X is given by:

P(X=k)=(kn​)p^k(1−p)^n−k

where n is the number of trials (7 in this case), k is the number of successes, and p is the probability of success on a single trial (0.8 in this case).

So, the probability that fewer than three of the selected adults watch prime-time TV live is:

P(X<3) = P(X=0) + P(X=1) + P(X=2)

=(7 0)(0.8)^0(0.2)^7 + (7 1​)(0.8)^1(0.2)^6 + (7 2​)(0.8)^2(0.2)^5

=1/78125 + 28/78125 + 336/78125

=73/15625

=0.004672

User AndreasPizsa
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