John and Peter are among the nine players a basketball coach

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

Probability approach:

1/9 choosing John (J);
1/8 choosing Peter (P);
7/7=1 choosing any (A) for the third player;
6/6=1 choosing any (A) for the fourth player;
5/5=1 choosing any (A) for the fifth player.

But scenario JPAAA can occur in \(\frac{5!}{3!}=20\) # of ways (JAPAA, JAAPA, AAAPJ, ... basically the # of permutations of the letters JPAAA, which is \(\frac{5!}{3!}=20\)).

So \(P=\frac{5!}{3!}*\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{5}{18}\).

Combinatorics approach:

P=favorable outcomes/total # of outcomes --> \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\).

\(C^2_2=1\) - # of ways to choose Peter and John out of Peter and John, basically 1 way;
\(C^3_7=35\) - # of ways to choose 3 other players out of 7 players left (without Peter and John);
\(C^5_9=126\) - total # of ways to choose 5 players out of 9.

Answer: D.

Hope it helps.
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Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

So according to the above # of permutations of 5 letters JPAAA would be 5!/3!, as there are 5 letters out of which A is represented thrice.

Hope it's clear.
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Total number of possibilities 9c5
Now say that John and Peter are already in the team. How can we pick rest of the three players out of 7 players - 7C3

Hence 7c3/9c5 = 5/18

Thanks a lot.

Since the order doesn't matter isnt James A Jacob AA or A James Jacob A A, same?
When do we need to use the multiplication factor (here 20) ingeneral? Also What does probability (1/9 * 1/8 * 3/7) represent

Order does not matter -YES, but favorable scenario JPAAA (John-Peter-Any-Any-Any) can occur in 5!/3! # of ways:
JPAAA (J - first, P second, any for the third, ... ) the probability of this particular scenario would be: \(P=\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{1}{72}\);

JAPAA (J - first, any (but P) for the second, P - third, ...) the probability of this particular scenario would be: \(P=\frac{1}{9}*\frac{7}{8}*\frac{1}{7}*1*1=\frac{1}{72}\);
...
...
...

There can be total of 20 such scenarios (20 as # of permuations of JPAAA is \(\frac{5!}{3!}=20\)) and each will have the probability of \(\frac{1}{72}\).

The final probability would be the sum of all these favorable scenarios: \(20*\frac{1}{72}=\frac{5}{18}\).

Hope it's clear.

You can check the Probability and Combination chapters of the Math Book (link below) for more.

Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event
probability-problem-90124.html#p684229
probability-85523.html#p641151
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Please explain why the permutation of jpaaa is not 5!.

Posted from my mobile device

Still a bit confused don't all the a's represent distinct items as they are different ways to write jpaaa as is jpa1a2a3 is not the same as jpa3a2a1

Posted from my mobile device

Not so. We are counting favorable scenarios to choose J and P. {J}{P}{A1}{A2}{A3} and {J}{P}{A3}{A2}{A1} represent one scenario: we choose J first, P second, any for third, any for fourth and any for fifth.

Check the links in my previous post for several similar problems to practice. Also if you are not comfortable with probability approach you can always use combinatorics method. Luckily probability questions can be solved in many ways, choose the one which you are most comfortable with.
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Dear Bunuel

Can we do this by 1-prob. of john and jack together.
I tried it but the result is not coming out right somehow.

This would be lengthier solution but yes we can:

P = 1 - (P(team with John and not Peter) + P(team with Peter but not John) + P(team without any of them)).
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I did the lengthier solution, but I get something different.

1-[Pr(Neither) + Pr(J, not P) + Pr(P, not J)]

1 - [3/18+ 2/18]

13/18

What did I do wrong?

Ex: Pr(Neither) = 7/9*6/8...*3/5 = 1/6

P(Both not selected) = (4/9) * (3/8) = 1/6 (The probability of John being among the 4 leftover out of 9 people is 4/9 and probability of Peter being among the leftover 3 is 3/8)
P(Peter selected, John is not) = (5/9)*(4/8) = 5/18 (Probability of Peter being among 5 selected is 5/9 and John being among the leftover 4 is 4/8)
P(Peter not selected, John is selected) = (5/9)*(4/8) = 5/18 (Probability of John being among 5 selected is 5/9 and Peter being among the leftover 4 is 4/8)

1 - (1/6 + 5/18 + 5/18) = 5/18

Hi Karishma,

Thank you for your response. There are still 2 things I do not understand.

The first is the fractions you use. Are we taking as given that the first 3 spots are filled? So for Pr(Neither), we're only considering the last spot, and we want to ignore J and P?

Second, where is the flaw in the method I used?

Pr(Neither) = 7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6
Pr(J, not P) = Pr(P, not J) = 1/9 * 7/8 * 6/7 * 5/6 * 4/5 = 1/18

Pr(J) is 1/9. Then we are only considering 7 of the remaining 8 people, then 6 of 7, etc.

Thank you.

The probability of selecting J is 5/9 since 5 people have to be selected and he can be any one of the 5. If you use this, you will get the correct answer. Though this is a very round about way of solving the question. Check out the solutions given on page 1.

P = 1 - (P(team with John and not Peter) + P(team with Peter but not John) + P(team without any of them)).

\(P = 1 - (\frac{1}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{5!}{4!} + \frac{1}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{5!}{4!} + \frac{7}{9}*\frac{6}{8}*\frac{5}{7}*\frac{4}{6}*\frac{3}{5})=\frac{5}{18}\).

We need to multiply first two cases by 5!/4! because JAAAA (John, Any but Peter, Any but Peter, Any but Peter, Any but Peter) and PAAAA (Peter, Any but John, Any but John, Any but John, Any but John) can occur in 5 ways.

Hope it helps.
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The number of ways of choosing 5 players from 9 is 9C5 = 9!/[5!(9 - 5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 )/( 4 x 3 x 2) = 3 x 7 x 6 = 126.

If John and Peter are already chosen for the team, then only 3 additional players must be chosen for the five-player team. The number of ways of choosing the 3 additional players from the remaining 7 players is 7C3 = 7!/[3!(7 - 3)!] = 7!/(3!4!) = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35.

Thus the probability that John and Peter will be chosen for the team is 35/126 = 5/18.

Answer: D
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VeritasKarishma :

Karishma : Why we use John selected as 5/9 and not 1/9 in case 2. Same goes for Peter.

When we use 5/9 as selection and when 1/9 for say "John".
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Probability of John being picked up first in the group of 5 = (1/9)*1*1*1*1
Probability of John being picked up second in the group of 5 = 1*(1/9)*1*1*1
and so on..

We add all these to get the probability of John being picked up in the group of 5 in whichever spot to get 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 5/9

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