What is the probability that the photograph shows a woman in the middle flanked by two men, one on either side of her?

Question

From a set of three women and four men, three are chosen at random to stand in a row for a photograph. What is the probability that the photograph shows a woman in the middle flanked by two men, one on either side of her?

my attempts:probability that the photograph shows a woman in the middle flanked by two men, one on either side of her $= \frac{2}{7}$.

personally I don't know this from where I have to start. I'm completely struck.

please help me.

Answer 1

Split it up into different problems. Firstly: exactly 2 men and exactly one woman must be chosen...so choosing 1 woman from a total of 3 is: $3 \choose{1}$ multiplied by choosing 2 men out of a total of 4: $4 \choose 2$.

Now the second part of the problem is this: after choosing the 3 people, how can they be arranged?: MWM, MMW, WMM, etc...since the one chosen woman MUST be in the middle, only the men on the sides can be arranged..so there are 2 possible "correct" arrangements, since it doesn't make a difference which of the two men is standing to her right.

So the total number of "correct" possibilities is: $${3\choose{1}}\cdot{4\choose{2}}\cdot2$$

All that is left is to divide that with the total number of possibilities to get the probability.

Answer 2

First of all, we need to choose $2$ men and $1$ woman to achieve this goal. There are $\binom{4}{2}\binom{3}{1} = 18$ ways to do this. Now, selecting them as this, we should find in how many ways they can be photographed with woman in middle. This one is easy because there are only $2$ ways (woman in middle, men can switch). So there are $18 \cdot 2 = 36$ ways satisfying the condition given.

Now, there are $\binom{7}{3}\cdot3!$ ways of choosing $3$ people and taking a photograph by switching them with $3!$ ways in total, which makes $210$ ways. Therefore the probability should be $\frac{36}{210} = \frac{6}{35}$.

Answer 3

Think of it like this: first the utmost left is chosen, then the utmost right and then the one for in between.

Then the answer is: $$\frac47\times\frac36\times\frac35$$

The first factor speaks for itself. The second is the probability that the utmost right will be a man under the condition that the first person chosen is a man. The third is the probability that the one in between will be a woman under condition that the first two persons chosen are men.