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Sagot :
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:
[tex]P(X=x)=\frac{n!}{(n-x)!\cdot x!}\cdot p^x\cdot(1-p)^{n-x}[/tex]From the problem, we identify:
[tex]\begin{gathered} n=22 \\ p=0.28 \end{gathered}[/tex]Then:
[tex]P(X=x)=\frac{22!}{(22-x)!\cdot x!}\cdot0.28^x\cdot0.72^{22-x}...(1)[/tex]Using the definition of the complement of an event, the probability that more than 7 will say yes is equivalent to:
[tex]P(X\gt7)=1-P(X\leqslant7)...(2)[/tex]Where:
[tex]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)[/tex]Using (1), we find each probability:
[tex]P(X=0)=\frac{22!}{(22-0)!\cdot0!}\cdot0.28^0\cdot0.72^{^{22-0}}=0.000726633[/tex][tex]P(X=1)=\frac{22!}{(22-1)!\cdot1!}\cdot0.28^1\cdot0.72^{^{22-1}}=0.00621675[/tex][tex]P(X=2)=\frac{22!}{(22-2)!\cdot2!}\cdot0.28^2\cdot0.72^{^{22-2}}=0.025385[/tex][tex]P(X=3)=\frac{22!}{(22-3)!\cdot3!}\cdot0.28^3\cdot0.72^{^{22-3}}=0.0658131[/tex][tex]P(X=4)=\frac{22!}{(22-4)!\cdot4!}\cdot0.28^4\cdot0.72^{^{22-4}}=0.121571[/tex][tex]P(X=5)=\frac{22!}{(22-5)!\cdot5!}\cdot0.28^5\cdot0.72^{^{22-5}}=0.1702[/tex][tex]P(X=6)=\frac{22!}{(22-6)!\cdot6!}\cdot0.28^6\cdot0.72^{^{22-6}}=0.187535[/tex][tex]P(X=7)=\frac{22!}{(22-7)!\cdot7!}\cdot0.28^7\cdot0.72^{^{22-7}}=0.166698[/tex]Taking the sum of all of them, and using (2):
[tex]\begin{gathered} P(X>7)=1-0.744146 \\ \\ \therefore P(X>7)=0.255854 \end{gathered}[/tex]
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