Christmas MT's answer to kristy's Junior College 2 H2 Maths Singapore question.
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Answer should be (5girls tgt and at least 4 boys tgt)/(at least 4 boys tgt)
P (5 girls tgt and at least 4 boys tgt)
= P (5 girls tgt and 4 bots tgt) + P (5 girls tgt and 5 boys tgt)
= (5!*5C4*4!*2!)/10! + (5!*5!*2!)/10!
= 1/63
P (at least 4 boys tgt)
= P (4 boys tgt) + P (5 boys tgt)
= (6!*5C4*4!*5)/10! + (6!*5!)/10!
= 1/7
Ans = (1/63) / (1/7) = 1/9
P (5 girls tgt and at least 4 boys tgt)
= P (5 girls tgt and 4 bots tgt) + P (5 girls tgt and 5 boys tgt)
= (5!*5C4*4!*2!)/10! + (5!*5!*2!)/10!
= 1/63
P (at least 4 boys tgt)
= P (4 boys tgt) + P (5 boys tgt)
= (6!*5C4*4!*5)/10! + (6!*5!)/10!
= 1/7
Ans = (1/63) / (1/7) = 1/9
Date Posted:
2 years ago
could you explain how you got P(4 boys tgt) ?
You group the four boys into one group, so the four boys and five girls give u 6 groups, hence the 6!.
You then need to choose four boys out of the five and can rearrange the four boys in any order so its 5C4 and 4!.
Lastly the remaining guy have 5 slots to stand such that he will not be beside any of the guys group so multiply by 5.
You then need to choose four boys out of the five and can rearrange the four boys in any order so its 5C4 and 4!.
Lastly the remaining guy have 5 slots to stand such that he will not be beside any of the guys group so multiply by 5.
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