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Bunuel
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Three quarters, three dimes, and x nickels are the only coins in a jar 07 May 2017, 14:21
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73% (01:29) correct 27% (01:24) wrong based on 310 sessions

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Three quarters, three dimes, and x nickels are the only coins in a jar. If six coins are drawn from the jar, what is the probability that the total value of the coins drawn is less than fifty cents?

(1) The probability of getting a total value of more than or equal to fifty cents is .83.
(2) There are 11 coins in the jar.
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D
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Three quarters, three dimes, and x nickels are the only coins in a jar 07 May 2017, 15:27
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Quarters: 3 * 0.25

Dimes: 3 * 0.10

Nickels: x * 0.01

Statement 1 : The probability of getting a total value of more than or equal to fifty cents is .83.

Probability that the total value of the coins drawn is less than fifty cents = 1 - 0.83 = 0.17

This statement is sufficient

Statement 2 : There are 11 coins in the jar.

There are already Three quarters and three dimes, so the remaining 5 coins are nickels.

As we know the total number of coins and the value of each coin, we can calculate the probability that the total value of the coins drawn is less than fifty cents.

This statement is sufficient.

Answer is D
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Three quarters, three dimes, and x nickels are the only coins in a jar 08 May 2017, 00:51
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stat 1: directly answers the question
Probability that the total value of the coins drawn is less than fifty cents = 1 - 0.83 = 0.17
suff

stat2: total 11 coins,,,so nickels = 5..since we have the individual coins value and numbers,, we can calculate,,
suff,,

ans D
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Three quarters, three dimes, and x nickels are the only coins in a jar 09 May 2017, 00:13
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Dimes, nickel.. I was not aware of the values of these.(these are the part of some currency i guess)?
In actual exam will I be provided with the values for those terms??
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Three quarters, three dimes, and x nickels are the only coins in a jar 09 May 2017, 02:21
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er.arun88
Dimes, nickel.. I was not aware of the values of these.(these are the part of some currency i guess)?
In actual exam will I be provided with the values for those terms??
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Yes, you are not required to know this. On the actual exam the values will be provided.
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Three quarters, three dimes, and x nickels are the only coins in a jar 11 May 2017, 08:15
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Statement 1 is quite clear: 1-0.83=0.17
Statement 2: Since we know all the values of coins and number of coins of each value, it is sufficient!
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Three quarters, three dimes, and x nickels are the only coins in a jar 11 May 2017, 21:53
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using statement 1, we can simply calculate the required probability. SUFFICIENT
Statement 2: since we know total no. of coins, and value of different types of coins,and also no. of coins of each category, it can be calculated. SUFFICIENT.

Hence, D is anwer!
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Three quarters, three dimes, and x nickels are the only coins in a jar 01 Jul 2017, 04:18
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Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.
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Three quarters, three dimes, and x nickels are the only coins in a jar 09 Jul 2017, 12:01
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TheMastermind
Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.
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Hi TheMastermind ,

I believe you understood from statement 2 that we have 5 coins of nickel.

Now, we have the following denomination.

3 coins of value $ 0.25 each.
3 coins of value $ 0.10 each.
5 coins of value $ 0.01 each.

Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25.

I cannot get $0.50 with any other combination such that only 6 coins are selected.

So, Probability of selecting atleast one quarter = 1 - P(selecting no quarter).

= 1 - 8C6/11C6.

Now, I know that this will give me value more than or equal to $0.50. So, 1- this probability will give me value less than $0.50.

I hope that makes sense. :)
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Three quarters, three dimes, and x nickels are the only coins in a jar 18 Oct 2018, 18:42
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abhimahna
TheMastermind
Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.

I believe you understood from statement 2 that we have 5 coins of nickel.

Now, we have the following denomination.

3 coins of value $ 0.25 each.
3 coins of value $ 0.10 each.
5 coins of value $ 0.01 each.

Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25.

I cannot get $0.50 with any other combination such that only 6 coins are selected.

So, Probability of selecting atleast one quarter = 1 - P(selecting no quarter).

= 1 - 8C6/11C6.

Now, I know that this will give me value more than or equal to $0.50. So, 1- this probability will give me value less than $0.50.

I hope that makes sense. :)
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Hi abhimahna,

regarding what you said,

"Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25."

if one coin selected from the value $0.25 and rest 5 from $0.01, then the value would not be >= $0.50

Could you please clarify this?

Thanks,
Shaheen
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Three quarters, three dimes, and x nickels are the only coins in a jar 08 Nov 2018, 01:04
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Hi Shaheen,

The value of a nickel is .05$ NOT .01$.
Now it makes sense to say that we've to select at least one coin of .25$
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Three quarters, three dimes, and x nickels are the only coins in a jar 28 Mar 2021, 10:00
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