Three men and 2 women will present 5 consecutive speeches

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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. 3124/3125
B. 9/10
C. 4/5
D. 16/25
E. 1/2

The OA is B.

Please, can anyone assist me with this PS question? How can I solve it? Thanks in advance!
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by GMATGuruNY » Thu May 24, 2018 3:54 pm
BTGmoderatorLU wrote: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. 3124/3125
B. 9/10
C. 4/5
D. 16/25
E. 1/2
P(at least 2 men speak consecutively) = 1 - P(no men speak consecutively).

For no men to speak consecutively, the order of the speeches must be as follows:
man-woman-man-woman-man.
P(1st speaker is a man) = 3/5. (Of the 5 speakers, 3 are men.)
P(2nd speaker is a woman) = 2/4. (Of the 4 remaining speakers, 2 are women.)
P(3rd speaker is a man) = 2/3. (Of the 3 remaining speakers, 2 are men.)
P(4th speaker is a woman) = 1/2. (Of the 2 remaining speakers, 1 is a woman.)
P(5th speaker is a man) = 1/1. (Only 1 speaker remains, and he is a man.)
To combine the probabilities above, we multiply:
P(no men speak consecutively) = 3/5 * 2/4 * 2/3 * 1/2 * 1/1 = 1/10.

Thus:
P(at least 2 men speak consecutively) = 1 - 1/10 = 9/10.

The correct answer is B.
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by [email protected] » Sun May 27, 2018 6:41 pm
Hi All,

We're told that 3 men and 2 women will present 5 consecutive speeches (1 by each person) at a conference and that the order of the speakers is determined randomly. We're asked for the probability that at least 2 of the men's speeches will be consecutive. This question can be solved in a couple of different ways, including by 'brute force' (simply 'drawing out' the possibilities).

With 3 men and 2 women, the following orders are possible:

MMMWW
MMWMW
MWMMW
WMMMW

MMWWM
MWMWM
WMMWM

MWWMM
WMWMM

WWMMM

There are just 10 possibilities (and they are all equally likely). Of these 10 options, 9 of the 10 include at least 2 consecutive speeches by men.

Final Answer: B

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by Scott@TargetTestPrep » Tue May 29, 2018 6:38 am
BTGmoderatorLU wrote: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. 3124/3125
B. 9/10
C. 4/5
D. 16/25
E. 1/2
Since there are 5 people, there are 5! = 120 ways the speeches can be presented. The only way the 2 of the men's speeches will not be consecutive is if the order is (M = a man's speech and W = a woman's speech): M-W-M-W-M. The number of ways this can be done is 3 x 2 x 2 x 1 x 1 = 12. So there must be 120 - 12 = 108 ways that at least two of the men's speeches will be consecutive and thus the probability is 108/120 = 9/10.

Alternate Solution:

Using the formula for permutations with indistinguishable objects, 3 men and 2 women can present speeches in 5!/(3! x 2!) = (5 x 4)/(2 x 1) = 10 ways.

Of these 10 orderings of speeches, only M-W-M-W-M does not meet the requirement of having at least two consecutive talks by men; thus, 10 - 1 = 9 orderings do meet this requirement. Thus, the probability is 9/10.

Answer: B

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