probability of buying candy

[mehrasa]

the prob that a visitor at the mall buys a pack of candy is .3.
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?


please give me your answers and explanations

p(buy candy)= 0.3 ==> p(not buy candy)=0.7
p(b,b,b,b,n,n)= (o.3^4)*(0.7^2)
there is a thing here, is the order of people who buy or not buy should be considered or not... i mean the # of arrangement
(bbbnbn)
(bbbnnb) and the like
if it is so the probability will be (0.3^4)* (0.7^2) * 6!/4!2!

[cans]

p(b) = 0.3 and p(n)=0.7
exactly4 = exactly 4 buy and 2 not buy..
select 4 persons from 6. 6C4 = 15
prob = 15* (0.3)^4*(0.7)^2 ( 4 buy and 2 not buy)

[aplavakarthik]

Bernoulli Trials, success=.3, failure =.7

(.3)^4*(.7)^2=0.003969

6c4*(.3)^4 * (.7)^2

order is not required

[knight247]

In this problem order IS important.
BBBBNN is not the same as BBNNBB. When you multiply with 6C4 which is 6!/4!2! you are actually multiplying with 6P6/4!2!=6!/4!2!. Which is a permutation with B repeated 4 times and N repeated twice.

@sandeep, don't read too much into the above statement as it can be a little confusing.
P(B)=0.3
P(N)=0.7

For four buys and two not buys we have (0.3)^4*(0.7)^2 and NNNNBB can be arranged in (6P6)/(4!2!) ways as N is repeated 4 times and B is repeated twice. (6P6)/(4!2!) =15 ways

So total number of ways is 15*(0.3)^4*(0.7)^2

[sl750]

6C4*(.3)^4*(.7)^2. This selects exactly 4 out of 6 people who buy candy with a probability of .3. It uses the Bernoulli trial formula

(p+q)^n = nC0*p^0*q^n + nC1*p^1*q^n-1 +...nCn-1*p^n-1*q^1+nCn*p^n*q^0
p - probability of success
q= 1-p
n- number of trials