Ten students, including Wilton and Katherine, are each entered into a

Question

Ten students, including Wilton and Katherine, are each entered into a raffle to win one of four bicycles. If each student holds one entry in the raffle, what is the probability that Katherine will win a bicycle but Wilton will not?
 
A. 4/35
B. 4/15
C. 2/5
D. 49/105
E. 1/2

ScottTargetTestPrep · Accepted Answer

We are given that 4 people will be selected from 10 people and need to determine the probability that, of the 4 people selected, Katherine is included but Wilton is not. Let’s first determine the total number of ways to select 4 people from a group of 10.
 
The number of ways to select 4 people from a group of 10 is:

10C4 = 10!/4!6! = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210

Next, let’s determine the number of ways to select 4 people from a group of 10 in which Katherine is included but Wilton is not.
 
Since Katherine must be included and Wilton is not, we reduce the number of available spots from 4 to 3 (because Katherine is already selected), and we reduce the number of people available to be selected from 10 to 8 (Wilton is not even considered, and Katherine is already selected).
 
Thus, the number of ways to select 4 people when Katherine is included and Wilton is not is:
 
8C3 = (8 x 7 x 6)/3! = (8 x 7 x 6)/(3 x 2 x 1) = 56 
 
So, finally, the probability that Katherine is selected but Witon is not is:
 
56/210 = 8/30 = 4/15 
 
Answer: B

broall · Answer

There are 4 bicycles for 4 winners. There are 10 participants, so the total selections is  C_{10}^4

If Katherine win and Wilton doesn't, there are 3 remaining bicycles for 8 remaining participants. So the selections is  C_{8}^3

The probability that Katherine will win a bicycle but Wilton will not is:  \frac{C_{8}^3}{C_{10}^4}=\frac{4}{15}

Shruti0805 · Answer

P(Katherine winning) =  \frac{4}{10} 
Total no. of candidates left after selecting Katherine = 9
So, P(Anybody else winning) =  \frac{3}{9}  => P(Anybody not winning) =  \frac{6}{9}

Total Probability =  \frac{4}{10} * \frac{6}{9}  =  \frac{4}{15}

Ghoskrp3 · Answer

Easiest way is by the method of Probability of dependent events. which is, P(E) of Katherine winning  multiplied  by the P(E) of Wilton not winning (or losing):

=>  Kathrine's winning P(E)  =   \frac{4C1}{10C1}  (meaning she is 1 of the 4 bicycle winners out of 1 of 10 participants)

=>   Wilton's  Losing P(E)  =   \frac{6C1}{9C1}  (meaning he is 1 of the 6 losers of the lot out of 1 of 9 participants, as we have just accounted Katherine's winning Probability and thus the competition is now in the remaining 9 participants)

Post Substitution it will give:  \frac{4C1}{10C1}  *  \frac{6C1}{9C1}

=>  \frac{4}{10}  *  \frac{6}{9}  =>  \frac{2}{5}  *  \frac{2}{3} 
 
 Hence B.  \frac{4}{15}

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