There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose

Question

There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chosen at random from the purse, what is the probability of getting more than 55 cents? 
(1 Dime = 10 Cents ; 1 Quarter = 25 Cents)

A. 1/5
B. 1/20
C. 1/2
D. 19/20
E. 4/5

LevanKhukhunashvili · Answer

Let's count the probability of getting 55 cents or less and then subtract the result from 1
We cant get less than 55 as there are only three 10 cent coins and the rest three are 25 cent-coins
We can get 55 cents if we pull one 25 cents and three 10 cents
dime, quarter, quarter, quarter: 3/6*3/5*2/4*1/3=1/20
dime, quarter, quarter, quarter --> can be arranged in 4!/3!=4 ways

probability of getting 55 cents is 1/20*4=1/5
probability of getting more than 55 cents is 1-1/5=4/5

IMO
Ans:  E

07tiger · Answer

Three Dimes: 10, 10, 10
Three Quarters: 25, 25, 25

Four Coins are picked.

So the minimum total we can get is 10+10+10+25 = 30+25 = 55
Which means any other total (combination) gives more than 55. Which means what is the probability of getting anything other than 55.

Well 55 is only one instance, so finding Probability for it is easy. So we are picking 4 coins.

Key- DL: Dimes Left, QL: Quarters Left, TCL: Total Coins Left 
 (DL/TCL)*(DL/TCL)*(DL/TCL)*(QL/TCL)*4 = (3/6)*(2/5)*(1/4)*(3/3)*4 = (1/20)*4 = 1/5

The probability of getting total 55 is 1/5 so the probability of getting anything but 55 is
= 1-(1/5) = (5-1)/5 =  4/5 (E)

KarishmaB · Answer

3 Dimes, 3 Quarters (3D, 3Q)

If we take only 1 quarter and all 3 dimes, we make exactly 55 cents. But we need more than 55 cents so we must take at least 2 quarters.

2 ways of making more than 55 cents:

Q, Q, D, D -> Probability = (3/6)*(3/5)*2 = 3/5 (Note that probability of picking 4 is the same as probability of discarding 2, a Q and a D)

Q, Q, Q, D -> Probability = (3/6)*(2/5) = 1/5 (The probability of picking these 3 is the same as the probability of discarding 2 Ds)

Total probability = 3/5 + 1/5 = 4/5

Answer (E)

Joseph67890 · Answer

2 combinations of dimes & quarters are possible
1 dime & 3 quarters >>3c1*3c3 = 3 combinations
2 dimes & 2 quarters >> 3c2*3c2 = 9 combinations
Total 12 combinations

Sample space = 6c4 = 15 combinations

probability = 12/15=4/5

Ans:E

TomNook · Answer

How I approached this question was to first think what ways could I get more than 55 cents with the 3 dimes and 3 quarters in the purse if 4 coins are randomly chosen.
You need at least 2 quarters = 0.25*2 = 0.5 and a dime to have more than 55 cents.

So the only way to have  NOT  more than 55 cents here is to have 1 quarter and 3 dimes because there are only 3 dimes and 3 quarters, you can't select 0 quarters. The order of the selection of the coins doesn't matter here.
To allocate 3 spots for dimes out of 3 dimes = 3C3
To allocate 1 spot for quarter out of 3 quarters = 3C1
For the denominator you are picking 4 coins from 6 total coins because there are 3 quarters and 3 dimes in the purse = 6C4

Knowing this I did P(more than 55 cents) = 1 -  \frac{(3C3 * 3C1)}{6C4} 
P(more than 55 cents) = 1 -  \frac{3}{15} 
P(more than 55 cents) =  \frac{4}{5}

Answer:  E

DanTheGMATMan · Answer

If you consider the possibilities, there are many ways to get more than 55 cents, but only one way to get exactly 55 cents (3 dimes and a quarter). And btw no ways of getting less than 55 cents. So, it's easier to do 1 minus the probability of getting exactly 55 cents.

Probability of getting 3 dimes and 1 quarter (choose 3 out of 3 dimes and 1 out of 3 quarters:

\frac{3!}{3!0!}  *  \frac{3!}{1!2!}  = 3

Choosing 3 coins out of 6 total:

\frac{6!}{3!3!}  =15

First value over the total value =  \frac{1}{5}

1-  \frac{1}{5}  =  \frac{4}{5}

There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose

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There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose

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