Total of 10 persons, out of which 6 are women.
So
probability of choosing the first woman = 6/10
probability of choosing the second woman = 5/9
probability of choosing the third woman = 4/8
Since we want all 3 steps to be a success, we need to have success in each of the steps, and the overall probability is given by the multiplication rule:
P(all 3 are women)=6/10*5/9*4/8=120/720=1/6
56/125-177/625= 280/625 - 177/625 = ( 280 - 177 ) / 625 = 103 / 625 = 0.1648
To find the total probability, we first need to solve for the probability of the individual events.
Event 1:
P(consonant)=7/11
We know this to be true because out of the 11 total possibilities for the event, 7 of them are consonants.
R E P L A C E M E N T
Event 2
P(e)=3/10
We know this to be true because out of the 10 total possibilities for the event (since we didn't replace the first card we withdrew), 3 of them are the letter 'e'.
R E P L A C E M E N T (-1 to account for the first card we withdrew)
The possibility of both events...
P(consonant then 'e')=7/11*3/10=21/110
Answer: P=21/110
By the binomial theorem,
![\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%28x%5E2-%5Cfrac1x%5Cright%29%5E6%20%3D%20%5Csum_%7Bk%3D0%7D%5E6%20%5Cbinom%206k%20%28x%5E2%29%5E%7B6-k%7D%20%5Cleft%28-%5Cfrac1x%5Cright%29%5Ek%20%3D%20%5Csum_%7Bk%3D0%7D%5E6%20%5Cbinom%206k%20%28-1%29%5Ek%20x%5E%7B12-3k%7D)
I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that
12 - 3k = 0 ===> 3k = 12 ===> k = 4
which corresponds to the constant coefficient
![\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}](https://tex.z-dn.net/?f=%5Cdbinom%2064%20%28-1%29%5E4%20%3D%20%5Cdfrac%7B6%21%7D%7B4%21%286-4%29%21%7D%20%3D%20%5Cboxed%7B15%7D)