yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
a. .343
b. .147
c. .189
d. .063
e. .027
Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.
Explanation:
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.
NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.
Let’s consider some similar examples:1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?
The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.
So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441
NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.
2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?At least ONE buys, means that buys exactly one OR exactly two OR exactly three:
P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657
P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)
P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)
P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.
BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:
P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.
3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?P(B=2)=5!/2!3!*0.3^2*0.7^3
We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.
Also discussed at:
https://gmatclub.com/forum/probability-8 ... ss#p641153Hope it helps.