A.) 7C1* 11C3/ 18C4
B.) 1 - (11C4/18C4)
C.) (11C4/18C4) + (7C1*11C3/18C4)
A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be
serviced on a particular Saturday morning. Because there are so few mechanics, only 6
can be serviced.
(a) If the 6 cars are chosen at random, what is the probability that 3 of the cars selected
are domestic and the other 3 are foreign?
(b) If the 6 cars are chosen at random, what is the probability that at most one domestic
car is selected?
(c) Does the binomial distribution apply to this problem? Explain why or why not?
Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains
A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman
Please help! Thank you.
B.) 1- (10C3/25C3)
These questions are not of Binomial theoram.
In binomial theory Q the probability is given. Elsewhere on this forum you might see Q like: On a given day probaility that it rains is 1/2. What is the probability that it will rain on exactly 4 out of 7 days.
Let me know if you have more questions on the same.
Probability = nCr * p^r * q^(n-r)
n = total outcome.
r = favorable outcome
p = probability of favorable outcomes
q = probability of unfavorable outcomes
in rain Question
p = 1/2
q = 1/2
n = 7
r = 4
7C4* (1/2)7 = 35/128
The question is not perfect. I just remember seeing it somewhere on the forum.
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hey Compboss, completely agree with you on all of your answers for Question1 and 2. I do have one question though in regards to Qustion 2 part C.Originally Posted by comp_bos
Can we answer the question like this?
The reason I ask is because I am not sure if I should multiply by 4 the different ways we can pick one women from 7. I multiplied by 4 because i thought that the woman can picked at 4 different times. please advise!
Also Cali, just to help you for your tes. Sometimes GMAT answers are in the form of nCr. Try and get some math course on this, but here is the real basi stuff needed to understand this.
C means combinations, this means that order does not matter.
So let's assume a question asks you the number of ways to choose 4 poeple from a group of 10. then you would write this as such: 10C4
The calculation for this is: n!/(r!(n-r)!) (I believe this is it)
So in our example it would be 10!/(4!(10-4)!)
Hope this helps a little. Just do a search ion these forums for Combinations. I am sure you will find a much better course on this.
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