(c)

Probablity that two buy candy is:

(0.3)(0.3)(0.7) = 0.063


Three ways that this can happen
A+B
A+C
B+C

3 * 0.063 = 0.189

Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy.

This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy.

(.3)(.3)(.7)*3 = .189
(.3)(.3)(.3)*1 = .027

.189 + .027 = .216 probability of at least 2 people buying candy.

0.3*0.3*0.7 + 0.3*0.3*0.7 + 0.3*0.3*0.7
= 0.189
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can someone explain why should we multiply by three and one,
thanks in advance

Since the three people are DISTINCT thts why the anser is .063x3=.189:)
_________________

countdown-beginshas-ended-85483-40.html#p649902

P(exactly 1 visitor buying candy) = 3 * 3/10*7/10*7/10 = .441 (Between A, B & C it could be A or B or C)

or, P(exactly 2 visitors buying candy) = 1 - 0.027 - 0.441 - 0.343 = .189

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

Soln:

Probability that exactly two will buy is = P(First Buys, Second Buys,Third does not buy) + P(First buys, Second does not buy,Third buys) + P(First does not buy, Second Buys,Third Buys)
= (3/10 * 3/10 * 7/10) * 3
= .189

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: probability-85523.html?hilit=certain%20junior%20class#p641153

Hope it helps.
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The binomial probability formula seems like overkill for this, but I like to use it when I can so I can remember how to use it...

This page explains the Binomial Probability formula: http://stattrek.com/lesson2/binomial.aspx


In this problem:
n=3
x=2
p=3/10

b(x;n,p) = 3C2 * (3/10)^2 * (7/10)
= {3 * 3 * 3 * 7} / 1000
= .189

An alternate approach:
S implies Success, F implies Failure
P(exactly two successes) = P (SSF) + P (SFS) + P (FSS)
= (3/10 * 3/10 * 7/10) + (3/10 * 7/10 * 3/10) + (7/10 * 3/10 * 3/10)
= 3 * (3 * 3 * 7 / 1000)
= .189