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exis [7]
4 years ago
7

Given a group of 8 women and 11 men, how many different ways are there of choosing one man and one woman for a committee?

Mathematics
1 answer:
Mashutka [201]4 years ago
8 0
Any of the 11 man can be chosen, and combined with any of the 8 women.

Assume we select man1. The selected committee can be:

(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8), 

so there are 8 committees selections with man1 in them.

we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.

so there are 11*8=88 ways of choosing 1 man and 1 woman.


Answer: 88
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We are given that a driver's education course compared 1,500 students who had not taken the course with 1,850 students who had.

Null Hypothesis, H_0 : p_1 = p_2 {means students who took the driver's education course and those who didn't took have same chances to pass the written driver's exam the first time}

Alternate Hypothesis, H_1 : p_1 > p_2 {means students who took the driver's education course were more likely to pass the written driver's exam the first time}

The test statistics used here will be two sample Binomial statistics i.e.;

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  Test Statistics = \frac{(\frac{1150}{1850} -\frac{1440}{1500})-0}{\sqrt{\frac{\frac{1150}{1850}(1- \frac{1150}{1850})}{1850} + \frac{\frac{1440}{1500}(1- \frac{1440}{1500})}{1500} } } = -27.38

P-value is given by, P(Z > -27.38) = 1 - P(Z > 27.38)

Now, in z table the highest critical value given is 4.4172 which corresponds to the probability value of 0.0005%. Since our test statistics is way higher than this so we can only say that the p-value for an appropriate hypothesis test is less than 0.999995 .

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