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2 Probability Problems

  • Thread starter Samurai44
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  • #1
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Homework Statement


Q1) In a group of 20 men and 12 women, half of the men have black eyes and a third of the women have black eyes. if a person is chosen randomly, what is the probability that the person is a women or black eyes?
Q2)
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The Attempt at a Solution


Q1)total people= 32 ,,, black eyes = 10/32 (men) + 4/32 (women) = 14/32
so a women OR black eyes = 12/32 + 14/32 = 26/32 .. BUT the correct final answer is 22/32 !
Q2) couldn't understand the question or solve it :(
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Q1) In a group of 20 men and 12 women, half of the men have black eyes and a third of the women have black eyes. if a person is chosen randomly, what is the probability that the person is a women or black eyes?
Q2)
View attachment 77144

The Attempt at a Solution


Q1)total people= 32 ,,, black eyes = 10/32 (men) + 4/32 (women) = 14/32
so a women OR black eyes = 12/32 + 14/32 = 26/32 .. BUT the correct final answer is 22/32 !
Q2) couldn't understand the question or solve it :(
For (1): use the inclusion-exclusion principle.

For (2): do you know what conditional probabililties are? You are being asked to calculate the probability that the cost is too low, given the person is in college.

Question (2) is poorly worded and very ungrammatical, and whoever wrote it should be ashamed of themselves. Also, the question does not make sense to me in practical terms: it is asking about the group of students who say "I am not paying enough for my college education".
 
  • #3
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Q1) In a group of 20 men and 12 women, half of the men have black eyes and a third of the women have black eyes. if a person is chosen randomly, what is the probability that the person is a women or black eyes?
Alternative approach for (1) to Ray Vickson's - how many of the people would count as a success in picking someone female or black-eyed?
(but I recommend you look up the inclusion-exclusion approach too... this is a nice case to understand it)

And I completely support RV on Q2 - the table is not well-described and there isn't even a question or a directive action verb like "State" or "Calculate". Here's my quick interpretation of what the text should say:
A survey was undertaken asking current and former students their opinion on the costs of attending college. The table below presents the proportion of respondents who answered in each of three possible categories, also split by whether they were in college presently or not.

Calculate the probability that a student in college would give the response that the cost of attending college is too low.
However, I can entirely believe that a population of students would have a proportion who respond that costs are too low, even if only to scramble the survey results :-)
 
  • #4
HallsofIvy
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Homework Statement


Q1) In a group of 20 men and 12 women, half of the men have black eyes and a third of the women have black eyes. if a person is chosen randomly, what is the probability that the person is a women or black eyes?
Q2)
View attachment 77144

The Attempt at a Solution


Q1)total people= 32 ,,, black eyes = 10/32 (men) + 4/32 (women) = 14/32
so a women OR black eyes = 12/32 + 14/32 = 26/32 .. BUT the correct final answer is 22/32 !
Q2) couldn't understand the question or solve it :(
Yes, there are 32 people, 12 of them women and 10 are men with black eyes. So a total of 12+ 10= 22 are women or have black eyes. Your first calculation counts all men and women with black eyes- You have the probability a randomly chosen person is a man or has black eyes.
Your second counts the 12 women and the 14 people, both men and women, who have black eyes- you are counting the women with black eyes twice!
 
  • #5
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For (1): use the inclusion-exclusion principle.
".
Alternative approach for (1) to Ray Vickson's - how many of the people would count as a success in picking someone female or black-eyed?
(but I recommend you look up the inclusion-exclusion approach too... this is a nice case to understand it)
Yes, there are 32 people, 12 of them women and 10 are men with black eyes. So a total of 12+ 10= 22 are women or have black eyes. Your first calculation counts all men and women with black eyes- You have the probability a randomly chosen person is a man or has black eyes.
Your second counts the 12 women and the 14 people, both men and women, who have black eyes- you are counting the women with black eyes twice!
Thanks you all guys!
So its like i have set "Black eyes" which equal to 14 , and set "women" which equal to 12 , and the intersection is 4 ..
so P(B) + P(W) - P(BnW) --> 14 + 12 - 4 =22
but how do i know when to use this principle ?
 
  • #6
haruspex
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how do i know when to use this principle ?
When counting members of overlapping sets.
 
  • #7
Ray Vickson
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Thanks you all guys!
So its like i have set "Black eyes" which equal to 14 , and set "women" which equal to 12 , and the intersection is 4 ..
so P(B) + P(W) - P(BnW) --> 14 + 12 - 4 =22
but how do i know when to use this principle ?
You use it whenever you need to find a probability involving "or", such as ##P(A\: \text{or}\:B)## and you know ##P(A), P(B), P(A \:\text{and} \: B)##. There are generalizations to three or more events, such as ##P(A \:\text{or} \: B \; \text{or} \: C)##, etc.
 
  • #8
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When counting members of overlapping sets.
You use it whenever you need to find a probability involving "or", such as ##P(A\: \text{or}\:B)## and you know ##P(A), P(B), P(A \:\text{and} \: B)##. There are generalizations to three or more events, such as ##P(A \:\text{or} \: B \; \text{or} \: C)##, etc.
That helps, thanks alot :D
 

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