Problem Solving

Hi, while answering the following question, I took the approach of using the 1-P(of picking a nickel on both occasions) so 1-(6/15*5/14) but it appears this is not a correct approach or alternative to the usual response to the below of 9/15*8/14.

Can you help explain why my first tried technique does not apply to this question?
Thanks

Question. In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?

Hi Orla,
This problem can be resolved through the direct approach as follows -

Selecting a coin other than nickel twice without replacing the previous is -
9/15*8/14 = 12/35

OA should be 12/35

Thanks, yes that approach I'm familiar with. Though I'm wondering why the problem cannot be solved using the indirect approach as mentioned in my post.

Any insight into this?

Hi Orla,

Refer this previous discussion on BTG.
https://www.beatthegmat.com/mgmat-cat-pr ... 53769.html

The opposite probability is two nickels in a row plus probabilities of selecting one nickel -
i.e. 6/15*5/14 + 6/15*9/14 + 6/15*9/14 (Here 6/15*9/14 is to be written twice because the selection of one nickel is possible in either first draw or second draw.

Thus, our opposite probability formula becomes 6/15*5/14 + 2*6/15*9/14 = 23/35
Thus, 1-P = 1-(23/35) = 12/35

This is the same as our reqd answer.

Hope this helped.

Orla M wrote:Hi, while answering the following question, I took the approach of using the 1-P(of picking a nickel on both occasions) so 1-(6/15*5/14) but it appears this is not a correct approach or alternative to the usual response to the below of 9/15*8/14.

Can you help explain why my first tried technique does not apply to this question?
Thanks

Question. In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?

The question stem is ambiguous.
Here, a coin other than a nickel is a penny or a dime.
Thus, the question stem is asking the following:
What is the probability of picking a penny or a dime twice in a row?

Twice in a row typically means the SAME OUTCOME in each case.
Thus, the wording here could imply that there are only two ways to get a favorable outcome:
penny - penny
dime - dime.

If the OA is [spoiler]12/35[/spoiler], then the INTENT of the question is to ask the following:
If two coins are selected without replacement, what is the probability that neither of the two coins selected is a nickel?

Solution:
P(first coin is not a nickel) = 9/15. (Of the 15 coins, 9 are not nickels.)
P(second coin is not a nickel) = 8/14. (Of the 14 remaining coins, 8 are not nickels.)
Since we want both events to happen, we multiply the fractions:
9/15 * 8/14 = [spoiler]12/35[/spoiler].

I took the approach of using the 1-P(of picking a nickel on both occasions) so 1-(6/15*5/14) but it appears this is not a correct approach or alternative to the usual response to the below of 9/15*8/14

P(good) = 1 - P(bad).

A bad outcome in this case is choosing AT LEAST ONE NICKEL.
Here are all of the ways to choose at least one nickel:
NN
NP
PN
ND
DN

Your solution accounts only for NN.
To determine the probability that NEITHER coin is a nickel, we must subtract ALL OF THE WAYS to choose at least one nickel:
P(neither coin is a nickel) = 1 - P(NN) - P(NP) - P(PN) - P(ND) - P(DN).

As you can see, it is more efficient to calculate directly the probability that NEITHER coin is a nickel, as shown in my solution above.