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probs/combos

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Senior | Next Rank: 100 Posts
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by maolivie » Mon Jul 09, 2007 10:30 am
Hey guys, these may be pretty easy, but the area I lack in is probabilities/combos - hence, the study time for it. Can you help me out with these questions? They are from a princeton book from the library, I wrote them down, but never wrote the answers for them. sorry in advance.

Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

What is the largest integer value of x that makes 30!/3^x an integer?

Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least one of the people chosen is a man?

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?
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by givemeanid » Mon Jul 09, 2007 10:45 am
maolivie, can you please split your questions into one per post? That way, it keeps the post focused and organized around one topic.

Eric, please correct me if I am mistaken in asking maolivie to do the above.


Thanks.
So It Goes
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by gabriel » Mon Jul 09, 2007 11:50 am
givemeanid wrote:maolivie, can you please split your questions into one per post? That way, it keeps the post focused and organized around one topic.

Eric, please correct me if I am mistaken in asking maolivie to do the above.


Thanks.
... actually the rule is to start a seperate thread for each question ....
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Re: probs/combos

by gabriel » Mon Jul 09, 2007 11:56 am
maolivie wrote:Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?
the probability of a tail is 1/2 .. now out of the five coins we have to choose 3 coins from the 5 coins we can do that in 5c3 ways .. now for these 3 coins the probability for tails will be 1/2*1/2*1/2 ... the other 2 coins will have heads side up and the probability of that happening is 1/2*1/2 .. so the total probability is 5c3*1/2*1/2*1/2*1/2*1/2
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Re: probs/combos

by gabriel » Mon Jul 09, 2007 12:06 pm
maolivie wrote:
What is the largest integer value of x that makes 30!/3^x an integer?
30! = 1*2*3*4*5*......*30 ... now for 30!/3^x to remain a integer and for x to have the maximum possible value, we will have to find the number of 3's in 30! ... that is we will have to find the power of 3 in 30! .... now, 3,6,12,15,21,24,30 each have 3^1 as a factor in them so the total number of 3's in these numbers is 7 because ... 9,18 have 3^2 as a factor so the number of 3's in these 2 numbers is 4 ... 27 has 3^3 as a factor so the number of 3's in it is 3 .. so the total number of 3's in 30! is 7+4+3 = 14 .. so the maximum value of x for 30!/3^x to remain a integer is 14 ...
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Re: probs/combos

by gabriel » Mon Jul 09, 2007 12:11 pm
maolivie wrote:Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least one of the people chosen is a man?
.. the group can be formed in 3 ways .. first.. 1 man and 2 women, this can be done in 4c1*6c2 .. second .. 2 men and 1 woman, this can be done in 4c2*6c1 .. third.. 3 men this can be done in 4c3 .. now without restrication the group can be chosen from the 10 in 10c3 ways .. so the probability is (4c1*6c2+4c2*6c1+4c3) / 10c3 ..
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Re: probs/combos

by gabriel » Mon Jul 09, 2007 12:16 pm
maolivie wrote:A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

this is somewhat similar to the 3rd question ... the necessary condition of having 2 officer and 2 civilians can be done in 5c2*9c2 ... now there is one place left in the committee this can be filled by either a civilian or a officer so thsi can be done in 10c1 ways( 10c1 bcoz after choosing 2 officers and 2 civilians there are 10 left in the group) .. so the total number of ways is 5c2*9c2*10c1 ..
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by maolivie » Mon Jul 09, 2007 12:43 pm
okay, just reposted all of them in there separate topic
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