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A committee is to be formed investigates what activities teenagers have available in small communities. The committee is to have 7, chosen randomly from interested community member. There are 10 parents, 5 teenagers, and 4 adults without children who have all expressed an interest in serving on the committee. What is the probability that the adults without children are all on the committee?
7 members
4 adults
19 people all together
7P4 = 840
19P4= 93024
7P4 /19P4
= .009--> 1%
Statistics Canada reported that approximately 76% of families in Edmonton consist of married couples. There are 22 students in grade 3. what is the probability that exactly half of the students have families where the parents are defined as married couples? What is the probability that at least one student does not have parents that are defined as married couples?
Number of trials: 22
Probability: 0.76
K= 11, 12, 13, 14, 15, 16
.. 23
Sum (binompdf (22, 0.76{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22})
= 0.998
Or do I do it this way:
I did it this way:
Number of trials: 22
Probability: 0.76
K= 1,2,3,4,5,6,7,8,9,10 11, 12, 13, 14, 15, 16
.. 23
Sum (binompdf (22, 0.76{ 1,2,3,4,5,6,7,8,9,10,11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22})
= 1
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For the first part, use P instead of C. The answer is
(Note: [sup]19[/sup]C[sub]7[/sub] is the number of ways of selecting 7 members from 19 people. [sup]15[/sup]C[sub]3[/sub] is for when you are selecting 3 people from 15 (that is to say, take the 4 childless adults out, and select 3 people from the remaining 15).)
The second part is a binomial problem. Probability that half the students have married parents is
To find probability that at least one of them has unmarried parents, find the opposite probability (i.e. none of them has unmarried parents which is to say, all of them have married parents). This is 0.76[sup]22[/sup]. So the answer you want is
Last edited by JaneFairfax (2007-06-05 08:21:23)
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wow thank you that makes more sense!:D:cool:
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