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by [email protected] » Thu Mar 10, 2016 4:28 pm
Hi didieravoaka,

When dealing with probability questions, there are only two results that can be calculated: what you WANT to have happen OR what you DON'T WANT to have happen. Those two outcomes create the following equation:

(Want) + (Don't Want) = 1

Sometimes it's actually easier to calculate what you WANT to have happen by calculating what you DON'T WANT to have happen (and then subtract that fraction from the number 1).

Here, we're asked for the probability of flipping AT LEAST one head and AT LEAST one tail on 6 coin flips. Since each coin flip has 2 possible outcomes, there are 2^6 = 64 possible outcomes (although there would be lots of 'duplicate results'). We don't want to have to determine every possible outcome that gives us at least 1 head and at least one tail though, so let's calculate the probability of that NOT happening.

There are two results that would NOT fit what we're looking for:

ALL HEADS
ALL TAILS

The probability of each is the same: 1/64

1/64 + 1/64 = 2/64 = 1/32

1/32 = about 3%

(Want) + (.03) = 1

Want = about 97%

Final Answer: E

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Rich
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by didieravoaka » Mon Mar 14, 2016 12:27 pm
Thanks Rich!

Can you detail the probability that NOT happening? Is that what gives 1/64?

Is that this way of thinking you apply... P(At least 1)=1-P(None)?

Thanks.
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by [email protected] » Mon Mar 14, 2016 9:41 pm
Hi didieravoaka,

When a probability question asks for something happening 'at least' a certain number of times, it's often easier to calculate what you DON'T WANT to have happen (in this case, the situations would NOT include at least one heads and one tails).

Here, the two options that would NOT fit what we're looking for are the "ALL HEADS" and "ALL TAILS" options. In all other situations, we'd have at least one heads and at least one tails. Since the "ALL HEADS" and "ALL TAILS" options are fairly easy to calculate, THAT method is arguably the fastest way to get to the correct answer.

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by Matt@VeritasPrep » Thu Mar 17, 2016 10:12 pm
didieravoaka wrote:Thanks Rich!

Can you detail the probability that NOT happening? Is that what gives 1/64?

Is that this way of thinking you apply... P(At least 1)=1-P(None)?

Thanks.
That equation is right, and it's a GREAT one to use in any problem that asks you for the probability of "at least one" thing happening: it tends to save a LOT of time.
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