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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  [#permalink]

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New post 09 Dec 2019, 07:16
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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
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Re: There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose  [#permalink]

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New post 09 Dec 2019, 08:41
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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
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Re: There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose  [#permalink]

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New post 09 Dec 2019, 19:28
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)
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Re: There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose  [#permalink]

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New post 09 Dec 2019, 22:43
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07tiger wrote:
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


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)
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There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose  [#permalink]

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New post 10 Dec 2019, 00:02
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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
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There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chose   [#permalink] 10 Dec 2019, 00:02
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