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Three men and 2 women will present 5 consecutive speeches, 1 by each person, at a conference...

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Three men and 2 women will present 5 consecutive speeches, 1 by each person, at a conference...

by AAPL » Fri Jul 10, 2020 4:10 am

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Three men and 2 women will present 5 consecutive speeches, 1 by each person, at a conference. If the order of the speakers is determined randomly, what is the probability that at least 2 of the men's speeches will be consecutive?

A. \(\dfrac{3124}{3125}\)

B. \(\dfrac{9}{10}\)

C. \(\dfrac{4}{5}\)

D. \(\dfrac{26}{25}\)

E. \(\dfrac{1}{2}\)

OA B
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each person, at a conference...

by deloitte247 » Sat Jul 11, 2020 7:23 am
What is the probability that at least 2 of the men's speeches will be consecutive?
$$u\sin g\ \frac{total\ cases\ -\ cases\ with\ no\ con\sec utive\ speeches}{total\ cases}$$
$$where\ total\ cases\ =\ \left(number\ of\ men\ +\ number\ of\ women\right)$$
$$=3+2!\ =\ 5!$$
Assuming a man's speech = m and woman's speech = w
The only way to have non-consecutive speeches from the men or women is if they are in the order m - w - m - w - m and the number of ways this can be done ;
=cases with no consecutive speeches = 3 * 2 * 2 * 1 * 1 =12
$$Hence,\ \frac{total\ cases\ -\ cases\ with\ no\ con\sec utive\ speeches}{total\ cases}$$
$$where\ total\ cases\ =\ 5!$$
$$=\ 5\cdot4\cdot3\cdot2\cdot1=120$$
$$cases\ with\ no\ con\sec utive\ speeches\ =\ 12$$
$$\frac{120-12}{120}=\frac{108}{120}$$
$$=\frac{9}{10}$$
$$Answer\ =\ B$$
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