SOLUTION: I really need help with these two questions. I am not sure if I have been overworking myself but, I don't know what to do anymore with them. Help!!! 1) On a Saturday evening, 34

Question 134527: I really need help with these two questions. I am not sure if I have been overworking myself but, I don't know what to do anymore with them. Help!!!
1) On a Saturday evening, 34% of the people in chicago go out to dinner, 18% see a movie, 13% have a party, and 35% stay home. If seven people are randomly selected, what is the probability that one eats out, three see a movie, two have a party and one stays home?
A) 0.099
B) 0.459
C) 0.102
D) 0.0049
2) A package of 10 batteries is checked to determine if there are any dead batteries. Four batteries are checked. If one or more are dead, the package is not sold. What is the probability that the package will not be sold if there are actually three dead batteries in the package?
A) 2/3
B) 6/7
C) 5/6
D) 4/5
Thank you for any help that you can give me. I am just not "getting" this stuff.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Comment: Again, these are multinomial probability problems.
The Formula is:
P = [n!/(x1!*x2!*x3!)]
n is the number of trials
x1,x2,x3 are the number of "successes" (a successes is whatever you are looking for.
p1 is the probability you are looking for x1 times
etc.
=================
1) On a Saturday evening, 34% of the people in chicago go out to dinner, 18% see a movie, 13% have a party, and 35% stay home. If seven people are randomly selected, what is the probability that one eats out, three see a movie, two have a party and one stays home?
P = [7!/(1!*3!*2!**1!)]
P = [420]
P = 0.0049..
---------------
Comment: Let me know if you do not understand this multinomial procedure.
-------------
A) 0.099
B) 0.459
C) 0.102
D) 0.0049
=======================================
2) A package of 10 batteries is checked to determine if there are any dead batteries. Four batteries are checked. If one or more are dead, the package is not sold. What is the probability that the package will not be sold if there are actually three dead batteries in the package?
---------
P(pick a bad battery) = 3/10 = 0.3
P(pick a good battery) = 0.7
----------------------
P(at least one bad in 4) = 1 - P(none bad in four)
= 1 - [7C4]/[10C4]
= 1 - (35/210)
= 0.83333333...
= 5/6

========================
Cheers,
Stan H.

A) 2/3
B) 6/7
C) 5/6
D) 4/5

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