Out of 5 boys and 6 girls, 4 students are sellected at rando

Question

Out of 5 boys and 6 girls, 4 students are selected at random, what is the probability that selection has more number of girls than the number of boys?

(A)  \frac{1}{33}

(B)  \frac{15}{33}

(C)  \frac{23}{66}

(D)  \frac{12}{33}

(E)  \frac{1}{3}

Princ · Accepted Answer

OA: C

Total Number of case of selecting four students out of 11  = C(11,4) = \frac{11!}{7!4!}=330

Number of ways of selecting 4 girls or 3 girls and 1 boy to form a 4 member team  = C(6,4)*1 + C(6,3)*C(5,1)=\frac{6!}{4!2!}*1 +\frac{6!}{3!3!}*\frac{5!}{4!1!}=115

Required probability  = \frac{115}{330} =\frac{23}{66}

EgmatQuantExpert · Answer

Solution

Given:  
 • Out of 5 boys and 6 girls, 4 students are selected at random.

To find:  
 • The probability that selection has more number of girls than the number of boys.

Approach and Working:  
From 11 students, 4 can be selected in  ^{11}C_4  = 330 ways

Now, if number of girls are more than boys, then the possible cases are:
 • 3 girls and 1 boy =  ^6C_3  x  ^5C_1  = 20 x 5 = 100
• 4 girls and 0 boy =  ^6C_4  = 15
• Hence, the probability =  \frac{100+15}{330} = \frac{115}{330} = \frac{23}{66}

Hence, the correct answer is option C.

Answer: C

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-AT-e1533292152363.png

CounterSniper · Answer

more number of girls than boys is possible in 2 cases

1. 3 girls and 1 boy =  \frac{6c3*5}{11c4} 
2. all 4 girls =  \frac{6c4}{11c4}

=  \frac{115}{330}  = \frac{23}{66}

sarusarthak · Answer

If we do it with the probability method, why is the answer different?

1st case : 3 girls and 1 boy 
6/11 x 5/10 x 4/9 x 5/8

2nd case : all are girls
6/11 x 5/10 x 4/9 x 3/8

And then we add both the cases, which is = 8/66

Where am I going wrong? Please suggest!

plaverbach · Answer

Total possibilities:  \frac{(11*10*9*8)}{(4*3*2)} =330
Only 1 boy:  \frac{(6*5*4*4)}{(3*2)} =80
Only girls:  \frac{(6*5*4*3)}{(4*3*2)} =15
80+15=95
 \frac{95}{330}=\frac{19}{66}

Can any one help me?

Ohhh
Only 1 boy:  \frac{(6*5*4* 5 )/(3*2)} =100

akadiyan · Answer

plaverbach  - You got the total possibilities and only girls count as 330 and 80 correct, but the only 1 boy should be 100 and not 80.

Total possibilities of selecting 4 students from 11 students =  ^{11}C_4  = 330

probability that selection has more number of girls than the number of boys :
Scenario # 1: 3 girls , 1 boy =  ^{6}C_3 * ^{5}C_1  = 100
Scenario # 2: 4 girls and 0 boys =  ^{6}C_4  = 15
possible outcomes = 100+15 = 115

Probability = 115/330 =  23/66

Ans:   C

plaverbach · Answer

Thanks  akadiyan , but how about  sarusarthak  solution?

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Out of 5 boys and 6 girls, 4 students are sellected at rando

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Out of 5 boys and 6 girls, 4 students are sellected at rando

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