Probability Q: Why doesn't 1-P(not happens) not work here?

[Orla M]

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?

[adthedaddy]

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

[Orla M]

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?

[adthedaddy]

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.

[GMATGuruNY]

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.