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probability help

by jzw » Fri Feb 24, 2012 1:00 pm
So there these two questions.

"A six sided die with faces numbered one through six is rolled twice. What is the probability that the face with the number 2 on it will NOT be facing upwards on either roll?"

So I took 5/6 x 5/6 and got 25/36. That's the correct answer. Fine. And now the next question...

"Bob flips a coin four times. What is the probablity that he gets heads on at least one of the four flips?"

So why here are we supposed to take the probability that it does happen (1/2 x 1/2 x 1/2 x 1/2 = 1/16) and subtract that from 1 leaving 15/16? I get that the probability something will happen is 1 minus the probability that it won't happen. But then, if we do that for the first problem (with the six sided die) then we would take 1/6 x 1/6 = 1/36 and then if subtract that from 1 we are left with 35/36 which is NOT the correct answer.

Simply put - I am not sure when we are supposed to take the probability of the opposite thing happening and subtract that from one, and when we're not. PLEASE HELP!! - WAIT - is the answer because of the words "at least"?
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by GMATGuruNY » Fri Feb 24, 2012 2:26 pm
jzw wrote:So there these two questions.

"A six sided die with faces numbered one through six is rolled twice. What is the probability that the face with the number 2 on it will NOT be facing upwards on either roll?"

So I took 5/6 x 5/6 and got 25/36. That's the correct answer. Fine. And now the next question...

"Bob flips a coin four times. What is the probablity that he gets heads on at least one of the four flips?"

So why here are we supposed to take the probability that it does happen (1/2 x 1/2 x 1/2 x 1/2 = 1/16) and subtract that from 1 leaving 15/16? I get that the probability something will happen is 1 minus the probability that it won't happen. But then, if we do that for the first problem (with the six sided die) then we would take 1/6 x 1/6 = 1/36 and then if subtract that from 1 we are left with 35/36 which is NOT the correct answer.

Simply put - I am not sure when we are supposed to take the probability of the opposite thing happening and subtract that from one, and when we're not. PLEASE HELP!! - WAIT - is the answer because of the words "at least"?
P(good) + P(bad) = 1.

Given any probability question, we can determine P(good) or 1-P(bad).
The result is the same.
We should use whichever approach is easier and/or faster.

In 2 rolls, P(not 2 on either roll):
Good outcome: not 2 on the first roll and not 2 on the second roll.
Bad outcomes: 2 only on the first roll, 2 only on the second roll, 2 on both rolls.
Since there is only one good outcome but several bad outcomes, P(good) is faster.

In 4 coin flips, P(at least one heads):
Good outcomes: 1 heads, 2 heads, 3 heads, or 4 heads.
Bad outcome: all tails.
Since there is only one bad outcome, 1-P(bad) is faster.
But then, if we do that for the first problem (with the six sided die) then we would take 1/6 x 1/6 = 1/36 and then if subtract that from 1 we are left with 35/36 which is NOT the correct answer.
Your calculations here don't account for all of the bad outcomes listed above:
P(2 only on the first roll) = 1/6 * 5/6 = 5/36.
P(2 only on the second roll) = 5/6 * 1/6 = 5/36.
P(2 on both rolls) = 1/6 * 1/6 = 1/36.
Thus, P(not 2 on either roll) = 1 - (5/26+5/36+1/6) = 25/36.
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by jzw » Fri Feb 24, 2012 5:02 pm
I'm not understanding how (conceptually) to figure it out the "longer" way in the six sided die question. Ie, if I was to change the question to "what are the chances the number 2 will land face up both times?" wouldn't the answer be 1/6 * 1/6 = 1/36 ?
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by GMATGuruNY » Fri Feb 24, 2012 5:20 pm
jzw wrote:I'm not understanding how (conceptually) to figure it out the "longer" way in the six sided die question. Ie, if I was to change the question to "what are the chances the number 2 will land face up both times?" wouldn't the answer be 1/6 * 1/6 = 1/36 ?
Step 1: What's a bad outcome?
NOT getting 2 both times.

Step 2: What are all of the ways to NOT get 2 BOTH times?
2 on the first roll but not on the second roll
2 on the second roll but not on the first roll
2 on neither roll.

Step 3: Are there more ways to get a good outcome or more ways to get a bad outcome?
In this case, there is only one way to get a good outcome (2 on each roll) but three ways to get a bad outcome, so it's quicker to determine P(good).

Here are other examples:

https://www.beatthegmat.com/combinations ... 02081.html

https://www.beatthegmat.com/baseball-t69721.html

https://www.beatthegmat.com/princeton-re ... 78922.html

https://www.beatthegmat.com/probability- ... 83693.html
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