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How to solve this Q?

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How to solve this Q?

by sanaa.rizwan » Mon Apr 08, 2013 4:38 pm
NP: OG PS 193

The probability is ½ that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails


A. 1/8
B. 1/2
C. 3/4
D. 7/8
E. 15/16
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by Anju@Gurome » Mon Apr 08, 2013 6:49 pm
sanaa.rizwan wrote:The probability is ½ that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails
Probability of having at least one tail = 1 - Probability of no tails = 1 - (1/2)*(1/2)*(1/2) = 1 - 1/8 = 7/8

The correct answer is D.
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by Brent@GMATPrepNow » Mon Apr 08, 2013 8:41 pm
sanaa.rizwan wrote:NP: OG PS 193

The probability is ½ that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails


A. 1/8
B. 1/2
C. 3/4
D. 7/8
E. 15/16
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)

Now let's calculate P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the first toss and heads on the second toss and heads on the third toss.
We can write P(getting zero tails) = P(heads on 1st AND heads on 2nd AND heads on 3rd)
This means that P(getting zero tails) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd)
Which means P(getting zero tails) = (1/2)x(1/2)x(1/2)= 1/8

We're now ready to answer the question.
P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
= 1 - 1/8
= [spoiler]7/8[/spoiler]
= D

Cheers,
Brent
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by vipulgoyal » Mon Apr 08, 2013 8:52 pm
Experts please suggest whats wrong with this approach

P of at least one tail

P of one tail + p of 2 tail + p of 3 tails

1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2

t * H * H + t * t * H + t * t * t

= 3/8
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by Brent@GMATPrepNow » Mon Apr 08, 2013 9:17 pm
vipulgoyal wrote:Experts please suggest whats wrong with this approach

P of at least one tail

P of one tail + p of 2 tail + p of 3 tails

1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2

t * H * H + t * t * H + t * t * t

= 3/8
You're missing a few scenarios.
For example, for 1 tails, you have THH, but what about HTH and HHT?

For your solution, we need:
P (at least 1 tails) = P(1 tails, or 2 tails or 3 tails)
= P(THH or HTH or HHT or TTH or THT or HTT or TTT)
= P(THH) + P(HTH) + P(HHT) + P(TTH) + P(THT) + P(HTT) + P(TTT)

Aside: Fortunately, each of these probabilities equals (1/2)(1/2)(1/2) = 1/8

= 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8
= 7/8
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