Question

According to government​ data, the probability that an adult was never in a museum is​ 15%. in a random survey of 10​ adults, what is the probability that two or fewer were never in a​ museum? round to the nearest thousandth.

Answer 1

Let p be the probability that an adult was never in a museum. Hence p = 0.15. Then q is the probability that an adult was in a museum is 1 - 0.15 = 0.75. We have a binomial expansion where the probability of k success in n trials is given by P_n(k) = (n, k) p^(k) q^(n -k) where (n, k) is the number of ways to select 10 objects from k things. At least two or fewer means we have P_10 (< or equal to 2) So we have P_10 (less than or equal to 2) = P_10 ( 0) + P_10 (1) + P_10 (2). So we have P _10 (0) = ( 10, 0) (0.15)^ (0) (0.75)^(0) = 0.196. For P_10(1), we have 0.3474 and for P_10(2), we have 0.2758. Adding these we have 0.1960 + 0.3474 + 0.2758 = 0.8192.

Answer 2

The probability that two or fewer were never in a​ museum is given by 0.8192.

Let p be the probability that an adult was never in a museum.

p = 0.15.

Then q is the probability that an adult was in a museum is

1 - 0.15 = 0.75.

We have a binomial expansion where the probability of k success in n trials is given by

What is the formula for binomial expansion?

$P_n(k) = (n, k) p^{(k)} q^{(n -k)}$

where (n, k) is the number of ways to select 10 objects from k things.

At least two or fewer means we have

$P_{10 }(\leq 2)$

So we have $P_{10}$ (less than or equal to 2) = $P_{10} ( 0) + P_{10} (1) + P_{10} (2).$

$P _{10 }(0) = ( 10, 0) (0.15)^ (0) (0.75)^(0) = 0.196.$

$For P_{10}(1)=0.3474$

$P_{10}(2)=0.2758.$

Adding these we have

0.1960 + 0.3474 + 0.2758 = 0.8192.

Therefore, The probability that two or fewer were never in a​ museum is given by 0.8192.

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