Question

Ten people, 3 women and 7 men, are running for 4 school board positions. If all
the candidates have an equal chance to win any of the 4 positions, what is the
probability that NO women will get a position?​

Answer 1

Answer:

yes the probality says no women get position

Answer 2

Answer:

The probability that no women will get a position is found to be 0.16667 or $\frac{1}{6}$.

Step-by-step explanation:

The total number of board positions in the school are: 4

and the total number of people running for it are: 10 (7 men and 3 women)

So, the total number of ways the board positions will be divided:

$^{10}C_4=\frac{10!}{4!\times(10-4)!}$

$^{10}C_4=\frac{10!}{4!\times6!}$

Simplifying this, we can say:

$^{10}C_4=210$

Now, the possibility where no women is chosen will be where all four positions are won by men, which means the number of ways it could be done are:

$^{7}C_4=\frac{7!}{4!\times(7-4)!}$

$^{7}C_4=\frac{7!}{4!\times3!}$

or we can say:

$^{7}C_4=35$

So, the probability where no women is chosen for the position is:

$P(no\ women\ in\ position)=\frac{number\ of\ ways\ no\ women\ gets\ chosen}{number\ of\ total\ possibilities}$

Substituting the above results, we get:

$P(no\ women\ in\ position)=\frac{35}{210}$

or we can say:

$P(no\ women\ in\ position)=\frac{1}{6}\ or\ 0.1667$