OK... the two ways are

1) Cases when none are there..
Total ways = (7+3)(7+3-1)=10*9=90
ways when both are men, => 3*2=6
thus ways when atleast one is woman => 90-6=84
Thus probability that atleast one is woman = \(\frac{84}{90}=\frac{14}{15}\)

2) find each case when women are present
a) only one is present
choose one female = 7C1
choose one male = 3C1
total ways = 7C1*3C1=7*3=21
but the female can be at either of the two position => 21*2=42
b) both are women
first one can be any of 7 and next any of remaining 6, so 7*6=42

Total ways women are present = 42+42=84
probability = \(\frac{84}{90}=\frac{14}{15}\)


A

At least 1 woman is selected if women and women is selected or women and men is selected or men and women is selected.

total number of possible outcomes is \(10 * 9 = 90\)

w&w = \(7*6 = 42\)
w&m = \(7*3 = 21\)
m&w = \(3*7 = 21\)
therefore the probability that at least 1 woman is selected is \(\frac{84}{90}\)which reduces to \(\frac{14}{15}\)

We can solve this using combination.
1 - (All Men Probability).

All men probability = 3C2 / 10C2 = 3/45 = 1/15

At least 1 Woman probability = 1-(1/15) = 14/15

A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)

We can solve this by using counting methods or by applying probability rules.
Let's use probability rules

We want P(selecting at least 1 woman)

When it comes to probability questions involving "at least," it's best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(selecting at least 1 woman) = 1 - P(not selecting at least 1 woman)
What does it mean to not select at least 1 woman? It means selecting ZERO women.
So, we can write: P(selecting at least 1 woman) = 1 - P(selecting ZERO women)


P(selecting ZERO women) = P(selecting TWO men)
= P(selecting a man to be president AND selecting a man to be vice-president)
= P(selecting a man to be president) x P(selecting a man to be vice-president)
= 3/10 x 2/9
= 6/90
= 1/15
ASIDE: For the 1st selection, 3 of the 10 people are men.
Once we select a man for the 1st selection, there are 9 people remaining, and 2 of them are men.


So, P(selecting at least 1 woman) = 1 - 1/15
= 14/15

Answer: A

Cheers,
Brent

A very comprehensive explanation as always, Brent!
The answer should be 'A', I think you made a typo saying 'E'.

Thank you.

Fixed

Thanks LT2018!
I've edited my response.

Cheers,
Brent

There are 3M and 7W

Prob. (1W for at-least one position) = 1 - (No W for both the positions)
So, we can select 2M from 3 in 3C2 ways.

Therefore,
Required Prob. = 1 - [3C2 / 10C2] = 14/15

Hence, option A

MW + WW (if no member can hold two offices simultaneously - means repetition not allowed)

MW = 3/10 * 7/9 * 2! = 21/45

WW = 7/10 * 6/9 = 42/90

21/45 + 42/90 = 14/15