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Probability again.
Message posted by Donna (via 64.12.106.26) on October 30, 2001 at 7:32 PM (ET)
#1
A pool of potential jurors consists of 10 mena nd 15 women. If two DIFFERENT people are randomly selected from this pool, find the probability that they are both women.
#2.
Back to that casino question... if you bet on any single number in roulette, your probability of winning is 1/38. Assume that you bet on a single number in each of 100 consecuative spins.
A) Find teh mean number of wins.
B) Find teh standard deviation
C) Would it be unusual to not win once in the 100 trials? (Hint- look back at the other casino question.)
#3.
Nine percent of men and 0.25% of women cannot distinguish between the colors red and green. if ten men and four women are randomly selected for a study of traffic signal perceptions, find the following probabilities:
A) Exactly three men cannot distinguish between red and green.
B) At least one woman cannot distinguish between red and green.
READERS RESPOND:
(In chronological order. Most recent at the bottom.)
Re: Probability again.
Message posted by Phil (via 165.247.234.26) on October 30, 2001 at 11:22 PM (ET)
Hint: If the probabilty of success doesn't change from trial to trial (p=constant), it is "sampling with replacement" and you would use the binomial distribution. If you don't replace the subject after each trial, it is "sampling without replacement" and you would use the hypergeometric.
You have two hypergeometric situations and one binomial.
That should help.
And thanks! Now I have questions for my quizzes on Thursday!
Re: Probability again.
Message posted by Phil (via 165.247.234.26) on October 30, 2001 at 11:30 PM (ET)
Oops! Make that two binomial and one hypergeometric.
Re: Probability again.
Message posted by Donna (via 64.12.107.176) on October 31, 2001 at 11:29 AM (ET)
Phil- we never learned hyper geometric so I dont know a single thing of what yoru talking about. And does the word different mean without replacement? im so confused.
Re: Probability again.
Message posted by Tomi (via 154.32.142.1) on November 1, 2001 at 12:14 AM (ET)
OK - ignore the hypergeometric distribution and solve Q1 by using a tree diagram. Yes, two DIFFERENT people does make it without replacement.P(2 women) = P (first choice is a woman) * P (second choice is a woman) = 15/25 * 14/24
Alternatively you could count the number of possible combinations. Have you done nCr and nPr yet?
P (two women) = (no of ways of choosing two women)/(no of ways of choosing two people).
Since the two women are different, you use nCr instead of nPr.
P = (15C2)/(25C2)
(Should give same answer as other method, but would be a better method if you were talking about five women rather than two - don't need a large tree diagram)
If you knopw how to use the binomial distribution, the other two are easy.
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