Jun and Aaron are equally good at chess.

Question

Hi, I'm new to this site to forgive me if I violate any rules.

I need help with this probability question:

A committee of 4 is to be chosen from group of 10 students, If Bob and Rachel are among those being considered, what is the probability of each of these evets?
a. Both Bob and Rachel will be on the committee.
b. Neither Bob and Rachel will be on the committee.

Would appreciate it if any one of you can reply asap! Thank u

Kyala1Jameson · Answer

A committee of 4 is to be chosen from group of 10 students, If Bob and Rachel are among those being considered, what is the probability of each of these evets?
 a. Both Bob and Rachel will be on the committee.
 
 Bob and Rachel will be in the committee . 
SO out of 4, we have only 2 members to choose from remaining 8 members .
 so 2 members can be selected from 8 members in 8C2 ways i.e 8! / 2! * 6! = 28 ways

b. Neither Bob and Rachel will be on the committee.
  
we have only 4 members to choose from remaining 8 members , as Neither Bob and Rachel will be on the committee.
 so 4 members can be selected from 8 members in 8C4 ways i.e 8! / 4! * 4! = 5*6*7*8/1*2*3*4 = 70 ways

SojuGirl · Answer

Thank you so much! Do I need to do anything else with a? I think It's the number of ways for the committee left but how about the probability for bob and rachel?

Where is my thought process going wrong?

Kyala1Jameson · Answer

Yes to find probability you just need to divide the total sample space with favorable case . 
so

a. Both Bob and Rachel will be on the committee.
 
 Bob and Rachel will be in the committee . 
SO out of 4, we have only 2 members to choose from remaining 8 members .
 so 2 members can be selected from 8 members in 8C2 ways i.e 8! / 2! * 6! = 28 ways

Total sample space = 10C4 = 10!/4!*6! = 210 ways

Required probability = 28 /210 = 2/15

b. Neither Bob and Rachel will be on the committee.
  
we have only 4 members to choose from remaining 8 members , as Neither Bob and Rachel will be on the committee.
 so 4 members can be selected from 8 members in 8C4 ways i.e 8! / 4! * 4! = 5*6*7*8/1*2*3*4 = 70 ways
Total sample space = 10C4 = 10!/4!*6! = 210 ways

Required probability = 70 /210 = 1/3

SojuGirl · Answer

Thank you! I greatly appreciate it : )

SojuGirl · Answer

A man has 5 blue and 4 yellow balls in a bag. Without looking in the bag, he takes out a ball, replaces it in the bag and takes out another blue? What is the probability that he:

a. gets 2 blue balls?
b. gets a blue ball and then a yellow ball?
c. gets a yellow ball and then a blue ball?
d. gets only one blue ball?

Thank you in advance! I'd appreciate it if any one of you can reply asap! : D

sumitkrocks · Answer

a. gets 2 blue balls?

No. of ways of picking two blue balls in two draws= 5 x 5 =25 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 25/81

b. gets a blue ball and then a yellow ball?
No. of ways of picking one blue ball and then yellow ball in two draws= 5 x 4 =20 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 20/81

c. gets a yellow ball and then a blue ball?
No. of ways of picking one yellow ball and then a blue ball in two draws= 4 x 5 =20 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 20/81

d. gets only one blue ball?
Gets only one blue ball i.e either in the first draw or in the second draw, so BY or YB these are two cases 
favorable cases= 20+20 
Total cases= 81
Probability = 40/81

Hope it helps

SojuGirl · Answer

Jun and Aaron are equally good at chess. They play three games against each other. What is the probability that:
a. Jun wins all three games?
b. Jun wins only the first two games?
c. Jun wins only one game?

Thank you in advance! I'd appreciate it if any one of you can reply asap! : D

HoneyLemon · Answer

Just a advanced level of similar question https://gmatclub.com/forum/bill-has-a-s ... 57417.html Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

kntombat · Answer

Hey jinroayeee , welcome to GMAT Club. Unfortunately, your post doesn't follow the strict rules regarding the format required for posting a question. In future before posting any new questions, please go through this : https://gmatclub.com/forum/rules-for-po ... 33935.html and then post.

SojuGirl · Answer

Hi! I'd like to ask help about this probability question:

Jun and Aaron are equally good at chess. They play three games against each other. What is the probability that:

a. Jun wins all three games?
b. Jun wins only the first two games?
c. Jun wins only one game?

Would appreciate it if any one of you can reply asap! Thank u

sujoykrdatta · Answer

They are equally good, so each has a probability of winning equal to 1/2

Jun wins all 3 
= J wins 1st AND J wins 2nd AND J wins 3rd
= 1/2 x 1/2 x 1/2 = 1/8

Jun wins first two 
= J wins 1st AND J wins 2nd AND A wins 3rd
= 1/2 x 1/2 x 1/2 = 1/8

Jun wins only one game: 
There are 3 cases depending on WHICH game J wins: 
J wins 1st AND A wins 2nd AND A wins 3rd
OR 
A wins 1st AND J wins 2nd AND A wins 3rd
OR
A wins 1st AND A wins 2nd AND J wins 3rd

= (1/2 x 1/2 x 1/2)
+ (1/2 x 1/2 x 1/2)
+ (1/2 x 1/2 x 1/2)

= 3/8

Posted from my mobile device

Kinshook · Answer

Given: Jun and Aaron are equally good at chess. They play three games against each other. 
Asked: What is the probability that:

a. Jun wins all three games?
= (1/2)^3 = 1/8
b. Jun wins only the first two games?
(1/2)^2(1/2) = 1/8
c. Jun wins only one game?
= (1/2)(1/2)^2 + (1/2)(1/2)(1/2) + (1/2)^2(1/2) = 3/8

Iamprithwish · Answer

From the question stem we have, p(Jun winning)=P(J)=1/2=p(Aaron winning)= P(A)

Case 1:- Jun wins all three games= p(Jun wins 1st game)xp(Jun wins 2nd game)xp(Jun wins 3rd game)= 1/2x1/2x1/2=1/8
Case 2:- P(J)XP(A)XP(A)=1/8
Case 3:- 3C1X(P(J))XP(A)XP(A)=3/8

soumya69 · Answer

a. gets 2 blue balls?

No. of ways of picking two blue balls in two draws= 5 x 5 =25 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 25/81

b. gets a blue ball and then a yellow ball?
No. of ways of picking one blue ball and then yellow ball in two draws= 5 x 4 =20 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 20/81

c. gets a yellow ball and then a blue ball?
No. of ways of picking one yellow ball and then a blue ball in two draws= 4 x 5 =20 (As he replaces ball after first draw)
No. of ways of picking two balls = 9 x 9 =81
Probability= Favorable cases/Total cases= 20/81

d. gets only one blue ball?
Gets only one blue ball i.e either in the first draw or in the second draw, so BY or YB these are two cases
favorable cases= 20+20
Total cases= 81
Probability = 40/81

ramlala · Answer

Hi jinroayeee Please follow the rules available in the below link, before creating a new post related to a GMAT question. https://gmatclub.com/forum/rules-for-po ... 33935.html

SojuGirl · Answer

Probability question:

A box contains five orange and two green marbles. What is the probability that an orange marble is drawn? If the marble is orange and is not replaced, what is the probability that the next marble is also an orange?

SojuGirl · Answer

If a number is selected at random from each of the sets (1,2,3) and (3,4,5), what is the probability that:

a. the sum of the two numbers will be greater than 5?
b. the product of the numbers will be greater than 5?

CrackverbalGMAT · Answer

Solution:

Number of orange marbles = 5

Number of green marbles = 2

Total sample space = 5+2 = 7

Probability that an orange marble is drawn = 5/7

Without replacing, the space has now 6 marbles with 4 orange and 2 green marbles

Probability that an orange marble is drawn = 4/6 = 2/3

Devmitra Sen
GMAT SME

HoneyLemon · Answer

Total marble 7 ---- 5 O , 2 G

probability that an orange marble is drawn = 5 / 7

If the marble is orange and is not replaced, what is the probability that the next marble is also an orange = 4/6 .

But if the ques asks that what is the probability that both marbles will be orange selecting one at a time : 5/7 * 4/6 = 20/42

Jun and Aaron are equally good at chess.

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