Three men and 2 women will present 5 consecutive speeches

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

GMAT assassins aren't born, they're made,
Rich

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