Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is :-
(a) 1/2
(b) 1/3
(c) 2/3
(d) 3/4
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Answered by
18
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is :-
As there are 5 places (3 for boys and 2 for girls), so total no of ways = 5!.
Total no of favorable placement = 5*3!*2!
Because 5 favorable placements -
-BGBGB
- GBBGB
- BGGBB
- GBGBB
- GGBBB
As a boy can change its position within three boys and a gal can in two gals.
So probability = 5*3!*2!/5!
= 5*3*2*1*2*1/5*4*3*2*1
= 1/2.
Option a is correct one.
Anonymous:
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hello can you help me to solve problems on Permutation and Combination which i posted yesterday pls if you don't mind?
Great Answer
i posted it check it in my question box.
see my third question that no one answered that question
do you got it?
Answered by
4
Hey mate...
here's the answer...
The Possible Outcomes are:
1. G1G1 G2G2 B1B1 B2B2 B3B3
2. G1G1 B1B1 G2G2 B2B2 B3B3
3. G1G1 B1B1 B2B2 G3G3 B3B3
4. B1B1 G1G1 G2G2 B2B2 B3B3
5. B1B1 G1G1 B2B2 G2G2 B3B3
The Probability is...
1−(2/5+1/10)=1/2
Option A is correct..
Hope this helps❤
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