Coin probability

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Coin probability

by EricKryk » Thu Feb 27, 2014 2:36 pm
The probability is 1/2 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 Brent@GMATPrepNow » Thu Feb 27, 2014 2:52 pm
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

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by Patrick_GMATFix » Thu Feb 27, 2014 2:53 pm
Because it must always be the case that either something happens at least once or it never happens, Prob(at least 1) + Prob(never) = 1. In general, Prob(at least one) can be solved easily by finding 1 - Prob(never). In this case, 1-Prob(no tails). I go through the question in detail in the full solution below (taken from the GMATFix App).

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by GMATGuruNY » Thu Feb 27, 2014 5:35 pm
EricKryk wrote:The probability is 1/2 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
Slightly different approach:
Good ways = total ways - bad ways.

Total ways:
Number of options for the first toss = 2. (Heads or tails.)
Number of options for the second toss = 2. (Heads or tails.)
Number of options for the third toss = 2. (Heads or tails.)
To combine these options, we multiply:
2*2*2 = 8.

Bad ways:
Of the 8 ways to toss the coin, only 1 way will yield NO TAILS:
HHH.

Good ways:
Total ways - bad ways = 8-1 = 7.

Resulting probability:
(good ways)/(total ways) = 7/8.

The correct answer is D.

Another option is to WRITE OUT all of the ways to toss the coin 3 times:
HHH
HHT
HTH
HTT
TTT
TTH
THH
THT

Of the 8 ways above, the 7 options in red each include at least one tails.
Thus, P(at least one tails) = 7/8.
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by Brent@GMATPrepNow » Fri Feb 28, 2014 7:36 am
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
Another approach is to draw a TREE DIAGRAM that shows all of the possible outcomes.

Here's the 1st toss:
Image

Then add the 2nd toss:
Image

Then the 3rd:
Image

So, as you can see there are 8 possible outcomes:
Image


Now let's place a checkmark beside those outcomes that satisfy the condition of having AT LEAST one tails.
Image

As we can see, 7 of the 8 outcomes satisfy this condition. So, the probability is [spoiler]7/8[/spoiler]

Answer: D

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by Scott@TargetTestPrep » Sat Jul 18, 2015 6:48 pm
EricKryk wrote:The probability is 1/2 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
Solution:

We can consider the following two events that could occur for the 3 coin flips: Either the coin will land on tails zero times, or the coin will land on tails at least one time. Remember that the phrase "at least one time" means "one or more." We can see that these two events are mutually exclusive events in which the probabilities of the two events add up to 1. That is,

P(landing on tails at least 1 time) + P(landing on tails zero times) = 1

Thus, we can say:

P(landing on tails at least 1 time) = 1 - P(landing on tails zero times)


Since we are tossing the coin 3 times, the outcome of zero tails in 3 tosses is the same as getting heads on all 3 tosses. We can calculate the probability of zero tails in 3 tosses as the probability of 3 heads in 3 tosses:

½ x ½ x ½ =1/8

Plugging this into our formula we have:

P(landing on tails at least 1 time) = 1 - 1/8

P(landing on tails at least 1 time) = 7/8

Answer: D

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by nikhilgmat31 » Wed Jul 29, 2015 5:44 am
we can also do it another way

THH can occur in 3 ways 3 * 1/2 * 1/2 * 1/2 = 3/8
TTH can occur in 3 ways 3 * 1/2 * 1/2 * 1/2 = 3/8
HHH can occur in 1 way 1/2 * 1/2 * 1/2 = 1/8

adding = (3+3+1)/8 = 7/8 is answer.