A basketball coach will select the members of a five-players team from armog 9 players, including John and Peter. If the five players are chosen at random, What is the probability that the coach chooses a team that includes both John and Peter?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
Answer is D
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John and Peter
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(2C2 x 7C3)/9C5=5/18ST wrote:A basketball coach will select the members of a five-players team from armog 9 players, including John and Peter. If the five players are chosen at random, What is the probability that the coach chooses a team that includes both John and Peter?
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
Answer is D
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imhimanshu
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let me explain this:
total number of outcomes=9C5 i.e selecting five persons from 9 guys
number of favourable = 7C3 since jhn and peter are already included. we need to select 3 guys out of remaing 7. i.e 7C3
so probability= 7C3/9C5 =5/18 hence D
total number of outcomes=9C5 i.e selecting five persons from 9 guys
number of favourable = 7C3 since jhn and peter are already included. we need to select 3 guys out of remaing 7. i.e 7C3
so probability= 7C3/9C5 =5/18 hence D
Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:real2008 wrote:why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
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skprocks
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That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.dtweah wrote:The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:real2008 wrote:why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?
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skprocks
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That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.dtweah wrote:The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:real2008 wrote:why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?
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outreach
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opposite event will be: none of them is in the team + only one of them is in the team
skprocks wrote:That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.dtweah wrote:The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:real2008 wrote:why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?
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GMATGuruNY
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The approach you're using is:skprocks wrote:That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.dtweah wrote:The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:real2008 wrote:why 2C2?dtweah wrote:Sorry for delay.ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G
Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5
2/2=2C2=1 x 7C3/9C5
That is the concept behind it.
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?
P(good outcome) = 1 - P(bad outcome)
To use this approach, we have to account for ALL the different ways to get a bad outcome:
P(not both J and P) = P(not J and not P) + P(J or P and 4 others)
P(not J and not P) = 7c5/9c5 = 1/6
P(J or P and 4 others) = 2 * 7c4/9c5 = 5/9
P(not both J and P) = 1/6 + 5/9 = 13/18
So P(both J and P) = 1 - 13/18 = 5/18.
I'd use this approach only when P(bad outcome) is easier to determine than P(good outcome). In this problem, P(good outcome) involves less math.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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