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# John flips a coin 4 times. What is the probability that he gets heads

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John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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Updated on: 29 Dec 2014, 02:47
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73% (00:54) correct 27% (00:59) wrong based on 113 sessions

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John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

Originally posted by tiamo2015 on 28 Dec 2014, 18:11.
Last edited by Bunuel on 29 Dec 2014, 02:47, edited 1 time in total.
Renamed the topic and edited the question.
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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28 Dec 2014, 20:25
1
Yes, E is correct.
The probability of getting a heads at least once = 1 - probability of getting tails all 4 times.
Ans = 1 - (1/2)(1/2)(1/2)(1/2) = 1 - (1/16) = 15/16
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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28 Dec 2014, 20:51
Total number of ways =2*2*2*2=16
Number of ways of not getting a single head..1
in all 16 possibilities there is only one in whcih we can get all tails
so probability of not getting a single head=1/16
now probability of getting atleast one head is 1-probability of not getting a single head
=1-1/16=15/16
Hope this helps..
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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28 Dec 2014, 23:30
The "1-x" short cut above is best here, but you can verify the long way.

There are 16 possible combinations here. They break down as follows:

1H, 3T: 4 ways (heads first, second, third, or fourth)
2H, 2T: 6 ways (heads in 1&2, 1&3, 1&4, 2&3, 2&4, or 3&4)
3H, 1T: 4 ways (as in the first one, but this time tails first, second, third, or fourth)
4H, 0T: 1 way (HHHH)
0H, 1T: 1 way (TTTT)

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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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29 Dec 2014, 20:14
DmitryFarber wrote:
The "1-x" short cut above is best here, but you can verify the long way.

There are 16 possible combinations here. They break down as follows:

1H, 3T: 4 ways (heads first, second, third, or fourth)
2H, 2T: 6 ways (heads in 1&2, 1&3, 1&4, 2&3, 2&4, or 3&4)
3H, 1T: 4 ways (as in the first one, but this time tails first, second, third, or fourth)
4H, 0T: 1 way (HHHH)
0H, 1T: 1 way (TTTT)

Thank you all for the answer and explanation, but I'm a bit confused as I've looked up few sources on the Internet (not relating to GMAT), and they all indicated the chance of getting 1 head in 4 flips of a coin is 4/16 = 1/4 , I'm quoting here from a site linked to University page:

Quote:
If we assume that each individual coin is equally likely to come up heads or tails, then each of the above 16 outcomes to 4 flips is equally likely. Each occurs a fraction one out of 16 times, or each has a probability of 1/16.

Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2)x(1/2)x(1/2)=(1/16).

Now lets ask: what is the probability that in 4 flips, one gets N heads, where N=0, 1, 2, 3, or 4. We can get this just by counting the number of outcomes above which have the desired number of heads, and dividing by the total number of possible outcomes, 16.

0 1 1/16 = 0.0625

1 4 4/16 = 1/4 = 0.25

2 6 6/16 = 3/8 = 0.375

3 4 4/16 = 1/4 = 0.25

4 1 1/16 = 0.0625

What is the difference here? is it the "at least" phrase?
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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29 Dec 2014, 21:30
What you're looking at there is the chance of getting heads exactly once, as opposed to at least once. That chance is represented by the first item in my list:

1H, 3T: 4 ways (heads first, second, third, or fourth)
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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08 Sep 2019, 05:52
tiamo2015 wrote:
John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

We can use the formula:

P(at least one head on 4 flips) = 1 - P(no heads on 4 flips)

P(no heads on 4 flips) = (1/2)^4 = 1/16.

P(at least one head on 4 flips) = 1 - 1/16 = 15/16.

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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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08 Sep 2019, 06:00
tiamo2015 wrote:
John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

Given: John flips a coin 4 times.

Asked: What is the probability that he gets heads on at least one of the four flips?

Probability of getting no heads on 4 flips = (1/2)^4 = 1/16

the probability that he gets heads on at least one of the four flips = 1 -1/16 = 15/16

IMO E
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Re: John flips a coin 4 times. What is the probability that he gets heads  [#permalink]

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08 Sep 2019, 06:19
The probability of having heads on at least one of the 4 flips = 1- the probability of having all tails in 4 flips.
= 1-(1÷16)
= 15/16.

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Re: John flips a coin 4 times. What is the probability that he gets heads   [#permalink] 08 Sep 2019, 06:19