help_me_learn
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1) in a particular area, 30% of men and 20% of women are underweight. four men and three women work in an office.
find the probability that there are 2 underweight people in the office.

2) on her drive to work stella has to go through four sets of traffic lights. she estimates that, for each set, the probability of her finding them red is 2/3 and green 1/3. (she ignores the possibility of them being amber) stella also estimates that when a set of lights is red she is delayed by one minute.
find the expected extra journey time due to waiting at lights.
for this question i did 4X(2/3)= 8/3
for that its 2.666 recurring however the answer is 2mins and 40 secs. where did the 40 sec come from?
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mqb2766
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(Original post by help_me_learn)
1) in a particular area, 30% of men and 20% of women are underweight. four men and three women work in an office.
find the probability that there are 2 underweight people in the office.

2) on her drive to work stella has to go through four sets of traffic lights. she estimates that, for each set, the probability of her finding them red is 2/3 and green 1/3. (she ignores the possibility of them being amber) stella also estimates that when a set of lights is red she is delayed by one minute.
find the expected extra journey time due to waiting at lights.
for this question i did 4X(2/3)= 8/3
for that its 2.666 recurring however the answer is 2mins and 40 secs. where did the 40 sec come from?
40 s is 2/3 of a minute.
What are you stuck with in Q1?
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help_me_learn
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(Original post by mqb2766)
40 s is 2/3 of a minute.
What are you stuck with in Q1?
finding the probability of two underweight people in the office.
i had initially tried to work out the probability of (MM) + (WW) + (MW) to add the probabilities together however that came out as wrong. i dont know how to do completely with that question.
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mqb2766
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(Original post by help_me_learn)
finding the probability of two underweight people in the office.
i had initially tried to work out the probability of (MM) + (WW) + (MW) to add the probabilities together however that came out as wrong. i dont know how to do completely with that question.
There are obviously two separate binomial groups to consider M & W.
Do them separately and combine in the manner you suggest.
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Aung
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Dude. Q1 is easy and binomial distribution is not necessary.

Just calculate underweight.
Men Men
Men Women
Women Men
Women Women

Like we did in elementary sch
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Aung
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(Original post by help_me_learn)
finding the probability of two underweight people in the office.
i had initially tried to work out the probability of (MM) + (WW) + (MW) to add the probabilities together however that came out as wrong. i dont know how to do completely with that question.

Multiply MW × 2
Becax MM + WW + MW + WM
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help_me_learn
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(Original post by mqb2766)
There are obviously two separate binomial groups to consider M & W.
Do them separately and combine in the manner you suggest.
i did
MEN: X~B (4, 30/100)
p(X=1) = 4C1 X (30/100)^1 X (70/100)^3 = 1029/2500
P(X=2)= 4C2 X (30/100)^2 X (70/100)^2 = 1323/5000
WOMEN: X~B (3,20/100)
P(X=1)= 3C1 X (20/100)^1 X (80/100)^2= 48/125
P(X=2)= 3C2 X (20/100)^2 X (80/100)^1= 12/125

OVERALL:
(WW) + (MM) +(MW)
12/125+ 1323/5000+ 2(1029/2500 X 48/125)
0.6767
Last edited by help_me_learn; 1 month ago
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help_me_learn
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(Original post by Aung)
Multiply MW × 2
Becax MM + WW + MW + WM
oh right i didnt think i needed to consider WM AND MW. thx
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help_me_learn
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#9
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(Original post by Aung)
Dude. Q1 is easy and binomial distribution is not necessary.

Just calculate underweight.
Men Men
Men Women
Women Men
Women Women

Like we did in elementary sch
you learnt these in elementary school? damn. im guessing thats basically primary school in the uk. we never learnt these until secondary aka middle school
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