Question

(CO 3) In the morning, about 28% of adults know what they will have for dinner. If one morning, you ask 22 adults if they know what they will have for dinner, what is the probability that more than 7 will say yes?

Answer 1

The probability that an adult knows what he will have for dinner: p = 0.28

We define event A as follows:

A: An adult knows what he will have for dinner

Then, if we ask 22 adults, the probability that more than 7 will say "yes" is given by the binomial distribution:

$P(X=x)=\frac{n!}{(n-x)!\cdot x!}\cdot p^x\cdot(1-p)^{n-x}$

From the problem, we identify:

$\begin{gathered} n=22 \\ p=0.28 \end{gathered}$

Then:

$P(X=x)=\frac{22!}{(22-x)!\cdot x!}\cdot0.28^x\cdot0.72^{22-x}...(1)$

Using the definition of the complement of an event, the probability that more than 7 will say yes is equivalent to:

$P(X\gt7)=1-P(X\leqslant7)...(2)$

Where:

$P(X\leqslant7)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)$

Using (1), we find each probability:

$P(X=0)=\frac{22!}{(22-0)!\cdot0!}\cdot0.28^0\cdot0.72^{^{22-0}}=0.000726633$$P(X=1)=\frac{22!}{(22-1)!\cdot1!}\cdot0.28^1\cdot0.72^{^{22-1}}=0.00621675$$P(X=2)=\frac{22!}{(22-2)!\cdot2!}\cdot0.28^2\cdot0.72^{^{22-2}}=0.025385$$P(X=3)=\frac{22!}{(22-3)!\cdot3!}\cdot0.28^3\cdot0.72^{^{22-3}}=0.0658131$$P(X=4)=\frac{22!}{(22-4)!\cdot4!}\cdot0.28^4\cdot0.72^{^{22-4}}=0.121571$$P(X=5)=\frac{22!}{(22-5)!\cdot5!}\cdot0.28^5\cdot0.72^{^{22-5}}=0.1702$$P(X=6)=\frac{22!}{(22-6)!\cdot6!}\cdot0.28^6\cdot0.72^{^{22-6}}=0.187535$$P(X=7)=\frac{22!}{(22-7)!\cdot7!}\cdot0.28^7\cdot0.72^{^{22-7}}=0.166698$

Taking the sum of all of them, and using (2):

$\begin{gathered} P(X>7)=1-0.744146 \\ \\ \therefore P(X>7)=0.255854 \end{gathered}$