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Find the probabilities using combinations and permutations. Leroy and Darnel made a list of 17 psychology experiments they want to try at home. Each experiment tackles one specific psychology topic. 13 of the experiments involve yawning. If Leroy and Darnel randomly choose 14 experiments to try over the summer, what is the probability that exactly 11 of the chosen psychology experiments involve yawning? Write your answer as a decimal rounded to four decimal places.



Answer :

The Solution.

The probability that an experiment involve yawning is given as below:

[tex]\text{Pr(Y)}=\frac{13}{17}[/tex]

The probability that an experiment does not involve yawning is

[tex]Pr(X)=1-Pr(Y)=1-\frac{13}{17}=\frac{4}{17}[/tex]

By Binomial expansion, we have that

[tex](Y+X)^{14}=^{14}C_0(Y^{14})(X^0)+^{14}C_1(Y^{13})(X^1)+\ldots^{14}C_{11}(Y^{11})(X^3)+\cdots[/tex]

So, the probability that exactly 11 out of the randomly chosen 14 experiments involve yawning is

[tex]Pr(\text{exactly 11 experiments involve yawning)=}^{14}C_{11}(Y^{11})(X^3)[/tex][tex]Pr(\text{exactly 11 experiments involve yawning)=}\frac{14!}{(14-11)!11!}_{}\times(\frac{13}{17}^{})^{11}(\frac{4}{17})^3[/tex][tex]Pr(\text{exactly 11 experiments involve yawning)=}\frac{14\times13\times12\times11!}{3!11!}_{}\times(\frac{13^{11}\times4^3}{17^{14}})^{}^{}^{}[/tex][tex]Pr(\text{exactly 11 experiments involve yawning)=}14\times13\times2_{}\times(\frac{13^{11}\times4^3}{17^{14}})[/tex][tex]\begin{gathered} Pr(\text{exactly 11 experiments involve yawning)= 364}_{}\times(\frac{13^{11}\times4^3}{17^{14}}) \\ \\ =364\times(\frac{13^{11}\times4^3}{17^{14}})=0.24796\approx0.2480 \end{gathered}[/tex]

Thus, the correct answer is 0.2480

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