Question #336618

An American Society of Investors survey found 30% of individual investors have used a discount broker. In a random sample of nine individuals, what is the probability:

a. Exactly two of the sampled individuals have used a discount broker?

b. Exactly four of them have used a discount broker?

c. None of them has used a discount broker? 


1
Expert's answer
2022-05-03T15:00:13-0400

We have a Bernoulli trial - exactly two possible outcomes, "success" (the individual investor has used a discount broker) and "failure" (the individual investor hasn't used a discount broker) and the probability of success is the same every time the experiment is conducted (an individual investor is examined), p=0.3,q=1−p=1−0.3=0.7, n=9.

The probability of each result

P(X=k)=(nk)pkqnk==(9k)0.3k0.79k==9!k!(9k)!0.3k0.79k.P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}9\\k\end{pmatrix}\cdot 0.3^k\cdot 0.7^{9-k}=\\ =\cfrac{9!}{k!\cdot(9-k)!}\cdot 0.3^k\cdot 0.7^{9-k}.


a. P(X=2)=9!2!7!0.320.77=0.2668.b. P(X=4)=9!4!5!0.340.75=0.1715.c. P(X=0)=9!0!9!0.300.79=0.0404.\text{a. } P(X=2)=\cfrac{9!}{2!\cdot7!}\cdot 0.3^{2}\cdot 0.7^{7}=0.2668.\\ \text{b. } P(X=4)=\cfrac{9!}{4!\cdot5!}\cdot 0.3^{4}\cdot 0.7^{5}=0.1715.\\ \text{c. } P(X=0)=\cfrac{9!}{0!\cdot9!}\cdot 0.3^{0}\cdot 0.7^{9}=0.0404.\\

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