SOLUTION: Help! A shipment of 11 televisions contains 4 regular and 7 deluxe models. The manufacturer failed to mark the model designation on the cartons. If 3 cartons are selected at ran

Algebra ->  Probability-and-statistics -> SOLUTION: Help! A shipment of 11 televisions contains 4 regular and 7 deluxe models. The manufacturer failed to mark the model designation on the cartons. If 3 cartons are selected at ran      Log On


   

Question 473930: Help!
A shipment of 11 televisions contains 4 regular and 7 deluxe models. The manufacturer failed to mark the model designation on the cartons. If 3 cartons are selected at random, what is the probability that exactly 2 of them are the deluxe model? (to three decimal places.)
Found 2 solutions by ccs2011, Maths68:
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Use Binomial Distribution
P+=+nCx+%2A+p%5Ex+%2A+%281-p%29%5E%28n-x%29
where n is total items, x is number of desired items, p is probability of selecting a desired item.
For this problem:
n = 3
x = 2
p = 7/11
P+=+3C2+%2A+%287%2F11%29%5E2+%2A+%284%2F11%29%5E1
P+=+3%2A%2849%2F121%29%2A%284%2F11%29
P+=+0.44177

Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!
Solution:-
Regular 1
Deluxe 2
Total 3
Now
Using nCr=n! /(n-r)! r!
3 things can be chosen out of 11 = 11C3 ways.
= 11!/(11-3)!3!
= 11!/ 8!3!
=(11*10*9*8!) / (8! * 3*2*1)
= (11*10*9) /(3*2*1)
= 990 /6
= 165 ways
1 Regular can be chosen out of 4 = 4C1
= 4! /(4-1)!1!
=4*3!/3!1!
= 4 ways
2 Delux can be chosen out of 7 = 7C2
=7!/(7-2)!2!
=7!/5!2!
=7*6*5!/5!2!
=7*6/2
=42/2
=21 ways
P(exactly 2 are deluxe model) = (7C2 * 4C1) / 11C3
= (21*4) /165
=84/165
=0.509