2 Probability/Combination Problems

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2 Probability/Combination Problems

by dpatwa » Thu Sep 06, 2007 8:58 am
Here are two question that I had trouble with, please provide some explanation:

Q#1
Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?
A 5/32
B 3/16
C 5/16
D 3/8
E 5/8

Q#2
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?

A 119
B 1,200
C 3,240
D 3,600
E 14,400

In the first question, I was able to obtain the answer by counting all combinations with 3 Tails, but I am having trouble finding a faster mathematical way of visualizing and solving the problem. The issue for me comes in determining the total number of combinations with 3 Tails facing up. In question two, again I could not figure out a straight forward way of dividing to remove the repeats. I would appreciate if someone could walk me through these. Thanks. Ans Q#1 C,Q#2 B

Source: Princeton Review Materials
Source: — Problem Solving |

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by samirpandeyit62 » Thu Sep 06, 2007 10:11 am
1st question

possible nos of outcomes when 1 coin is tossed = 2 H OR T

for 5 coins tossed is 2^ 5 =32

A coin can show tail in one way only

3 tails out of 5 can be obtained in 5C3 =10 ways

so probabilty is 10/32= 5/16

ans should be C

2nd q

case 1: committe has 3 offs & 2 civils

3 offs can be chosen out of 5 in 5C3 =10ways
2 civils can be chosen in 9C2 = 36 ways

hence total =10 * 36 =360 ways

case 2: 2 offs & 3 civls

hence offs can be selected in 5C2 =10 ways
civils can be selected in 9C3 =84 ways

total =10 * 84 =840 ways

either case can happen hence i.e case 1 or case2

hence total combinations = 360 +840 =1200

so ans should be B

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Re: 2 Probability/Combination Problems

by ri2007 » Thu Sep 06, 2007 10:17 am
dpatwa wrote:Here are two question that I had trouble with, please provide some explanation:

Q#1
Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?
A 5/32
B 3/16
C 5/16
D 3/8
E 5/8

Q1) I would use formula for binomial probability and the answer would be 5/16 or C

I got this page on the net to explain better you can check it out

https://faculty.vassar.edu/lowry/binomialX.html

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by dpatwa » Sat Sep 08, 2007 11:13 am
In Q#1, how does one come up with that there are 10 possible ways to obtain and 3/2 combination for tails and heads?

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by Bharat » Sun Sep 09, 2007 11:51 pm
Hi,
the possible ways to select 3 objects out of 5 is possible in 5C3 ways only.
5C3 = (5!)/(3!)((5-3)!) = 10

Let me know if this is not clear.