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03 Apr 2020, 22:31
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:46) correct
31%
(03:19)
wrong
based on 100
sessions
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A committee of 5 has to be formed from 4 Adults and 6 children. Find the probability that the committee has at least 2 children?
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
04 Apr 2020, 00:16
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Dillesh4096
A committee of 5 has to be formed from 4 Adults and 6 children. Find the probability that the committee has at least 2 children?
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
Two ways..
1) Find the opposite, that is where there are less than 2 children...
a) 1 children - 4 adults and 1 children = 4C4*6C1=1*6=6
b) NO children - Not possible as there are only 4 left then.
Total ways = \((4+6)C5=10C5=\frac{10!}{5!5!}=\frac{6*7*8*9*10}{1*2*3*4*5}=7*4*9\)
Probability of <2 children = \(\frac{4}{7*8*9}=\frac{1}{42}\)..
So, probability of at least 2 children = \(1-\frac{1}{42}=\frac{41}{42}\)
1) Find the ways where there are at least 2 children...
a) 2 children - 3 adults and 2 children = 4C3*6C2=4*15=60
b) 3 children - 2 adults and 3 children = 4C2*6C3=6*20=120
c) 4 children - 1 adult and 4 children = 4C1*6C4=4*15=60
d) 5 children - 0 adult and 5 children = 4C0*6C5=4*15=6
total ways with restriction of atleast 2 children = 60+120+60+6=246
Total ways = \((4+6)C5=10C5=\frac{10!}{5!5!}=\frac{6*7*8*9*10}{1*2*3*4*5}=7*4*9\)
Probability of at least 2 children = \(\frac{246}{7*8*9}=\frac{6*41}{42}=\frac{41}{42}\)
C
04 Apr 2020, 01:21
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Dillesh4096
A committee of 5 has to be formed from 4 Adults and 6 children. Find the probability that the committee has at least 2 children?
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
A. 31/42
B. 124/126
C. 41/42
D. 3/26
E. 11/42
Method-1:
Committe with at least 2 children could have following compositions
I) 2 children and 3 adults - \(6C_2*4C_3 = 15*4 = 60\)
II) 3 children and 2 adults - \(6C_3*4C_2 = 20*6 = 120\)
II) 4 children and 1 adult - \(6C_4*4C_1 = 15*4 = 60\)
IV) 5 children - - 6C_5 = 6
Total Favorable cases = 60+120+60+6 = 246
Total ways of choosing committee of 5 out of 10 members = \(10C_5 = 252\)
REQUIRED PROBABILITY = 246/252 = 123/126 = 41/42
Answer: Option C
Method-2:
Total - Unfavorable = Favourble \(= 1- [4C_1*6C_4/10C_5] = 41/42\)
Answer: Option C
