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junior college 2 | H2 Maths
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i am unable to get the ans for part (iv) the ans is 1/9. pls help thank you :)
Date Posted: 2 years ago
Views: 486
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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
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.
sorry could you explain further how to get P(Y) case 1 where 4 boys are tgt
P(Y) case 1 is: First arrange the 5 girls in 5! ways. Then the four of the five boys must be arranged in 5P4 ways as I only want four of them first. As the group of four-boys must not be next to the remaining 1 boy, I have to use the slot-in method.
Imagine, you space the five girls apart.
__ G1 __ G2 __ G3 __ G4 __ G5 __
Including the two slots at both ends, there are six empty slots. First I put in the group of 4-boys, there are 6 choices. Next, the remaining boy has 5 slots to choose from. So these are the 6*5 or 6P2 ways.
Imagine, you space the five girls apart.
__ G1 __ G2 __ G3 __ G4 __ G5 __
Including the two slots at both ends, there are six empty slots. First I put in the group of 4-boys, there are 6 choices. Next, the remaining boy has 5 slots to choose from. So these are the 6*5 or 6P2 ways.
oh nice! i understand now, thank you so much :D
No worry. I am gladed too!:)
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