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.
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