A hand purse contains 6 nickels, 5 pennies and 4 dimes. What is the pr

PENNIES (P) NICKEL(N) DIME(D)

Total = 15 coins
Nickel = 6
Not Nickel = 9
Probability of taking a coin other than Nickel = 9/15=3/5 ( first withdrawal)
If the coin taken is not replaced....
Probablility of taking a coin other than Nickel = 8/14= 4/7 (2nd withdrawal)

Total probability of drawing a coin other than Nickel in two successive withdrawals= 3/5 * 4/7= 12/35

Ans D

The probability of picking a coin that is not a nickel, twice in a row, is 9/15 * 8/14 = 3/5 * 4/7 = 12/35.

Answer: D

Solution

Given:

• A hand purse contains 6 nickels, 5 pennies and 4 dimes

To find:

• The probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back

Approach and Working Out:

• Total cases = \(^{15}C_1 * ^{14}C_1\)
• Total favorable cases =\( ^9C_1 * ^8C_1\)

Therefore, the required probability = \(9 * \frac{8}{15} * 14 = \frac{12}{35}\)

Hence, the correct answer is Option D.

Answer: D

Hi All,

We're told that a hand purse contains 6 nickels, 5 pennies and 4 dimes. We're asked for the probability of picking two coins OTHER than nickels if the first coin picked is NOT put back. This question is a straight-forward Probability question, so we just have to work step-by-step and do the necessary calculations...

There are 6+5+4 = 15 total coins.

The probability of NOT choosing a nickel on the first try is 9/15 = 3/5

Since we do NOT put that coin back, we have removed one non-nickel from the purse and the probability of NOT choosing a nickel on the second try is 8/14 = 4/7

Thus, the overall probability of pulling two NON-nickels is (3/5)(4/7) = 12/35

Final Answer:

GMAT assassins aren't born, they're made,
Rich

The P of getting Nickle in first attempt = 6/15
The P of getting Nickle in second attempt = 5/14

P( Nickle) = 6/15 *5/14 = 1/7

What am I doing wrong here ?

Bunuel

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We need the probability of picking a coin that is NOT a nickel, twice in a row. Please review the solutions above.