IBM
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Numerical Ability
Permutation and Combination
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A. 21/46
B. 25/117
C. 1/50
D. 3/25
Read Solution (Total 4)
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- The correct option is A.
as we have to choose combination of boys and girl
so required probability given by (15C2*10C1)/25C3
= ( (15*14)/2*10)/((25*24*23)/(3*2))
=21/46 - 5 years agoHelpfull: Yes(8) No(0)
- 10c1*15c2/25c3. On solving this u will get A is the ans
- 5 years agoHelpfull: Yes(0) No(1)
- Let , S - sample space E - event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= 25C325C3
= 2300.
n(E) = 10C1×15C210C1×15C2 = 1050.
P(E) = n(E)/n(s) = 1050/2300 = 21/46 - 5 years agoHelpfull: Yes(0) No(0)
- 25c3 = 2300
combination for the number 3 ways is 15*7*10 = 1050
the probability that 3 students come is = 1050/2300 = 21/46 - 4 years agoHelpfull: Yes(0) No(0)
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