# Thread: probability of choosing 4 person

1. ## probability of choosing 4 person

for part b , my working is (12c2 x 18c2)/(35c4) = 0.192

thers're 12 women doctor , 2 are chosen , there're 18 male left, 2 are chosen . no condition= choose 4 person from 35 person. but the ans is 0.807

2. ## Re: probability of choosing 4 person

How many possible panels have two women and two men? That would be $\displaystyle {_{17}C_2}\cdot {_{18}C_2}$. So, your denominator is wrong.

Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be $\displaystyle \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49$ I dunno where 0.807 comes from.

3. ## Re: probability of choosing 4 person

Originally Posted by SlipEternal
How many possible panels have two women and two men? That would be $\displaystyle {_{17}C_2}\cdot {_{18}C_2}$. So, your denominator is wrong.

Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be $\displaystyle \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49$ I dunno where 0.807 comes from.

the question ask for 4 person to be chosem in the panel disussion am i right? which include 2 woman doctor..

4. ## Re: probability of choosing 4 person

Originally Posted by SlipEternal
How many possible panels have two women and two men? That would be $\displaystyle {_{17}C_2}\cdot {_{18}C_2}$. So, your denominator is wrong.

Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be $\displaystyle \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49$ I dunno where 0.807 comes from.
the question ask for 4 person to be chosem in the panel disussion am i right? which include 2 woman doctor..

5. ## Re: probability of choosing 4 person

The problem said assuming there are exactly two women on the panel, what is the probability that both are women. This is exactly the same as asking "If two women are selected from a group of 17, 12 of whom are doctors, what is the probability they are both doctors. The number of men is not relevant.

6. ## Re: probability of choosing 4 person

Had the question been worded, "If at least two women are chosen for the panel, what is the probability that at least two women doctors are chosen", then the answer would be:

$\displaystyle \dfrac{ {_{12}C_2}\cdot {_{23}C_2} + {_{12}C_3}\cdot {_{23}C_1} + {_{12}C_4}\cdot {_{23}C_0} }{ {_{17}C_2}\cdot {_{18}C_2} + {_{17}C_3}\cdot {_{18}C_1} + {_{17}C_4}\cdot {_{18}C_0} } = \dfrac{1309}{2084} \approx 0.628$

I don't think I know of a question similar to the one asked that could yield a probability as high as they want.