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If the probability that an unfair coin displays heads after it’s flipp

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If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 15 Jun 2018, 23:17
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If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 07 Jul 2018, 01:21
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prags1989 wrote:
I still don't understand this question. Can someone help?


If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

A fair coin has the probability of heads equal to the probability of tails, so P(tail) = P(head) = 1/2. We are told that the coins is unfair so P(tail) ≠ P(head).

The probability that the coin displays heads after it’s flipped is three times the probability that it will display tails: P(h) = 3*P(t). Sin, the sum of these probabilities must be 1, then P(h) + P(t) = 1:

3*P(t) + P(t) = 1;

P(t) = 1/4 and P(h) = 3/4.

The question ask to find the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times. Exactly 3 heads can occur in several different ways:
HHHTT
HHTHT
HTHHT
THHHT
THHTH
THTHH
TTHHH
HHTTH
HTTHH
HTHTH
Basically, this is permutations of 5 letters HHHTT, out of which 3 H's and 2 T's are identical: 5!/(3!2!) = 10.

Now, each of the above 10 cases have the probability of (3/4)^3*(1/4)^2. Thus, the overall probability of P(HHHTT) is \(10*(\frac{3}{4})^3*(\frac{1}{4})^2 =\frac{135}{512}\).

Answer: B.

Hope it's clear.
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 30 Sep 2018, 08:09
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This problem can be solved by binomial probability formula.

Probability of doing a thing x times out of total n ways= \(nCx * (p^x) ( 1- p )^ (n –x )\)

where n denotes the number of trials and p denotes the success probability.

Required probability = \((5C3) * (3/4)^3 * (1/4)^2\)

= 135/512
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 05 Dec 2019, 19:17
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Bunuel wrote:
If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32



Since P(H) = 3P(T) and P(H) + P(T) = 1, then P(H) = ¾, and P(T) = ¼. Therefore, the probability of getting heads 3 times followed by tails 2 times is:


P(HHHTT) = (¾)^3 x (¼)^2 = 27/64 x 1/16 = 27/1024

However, since 3 heads and 2 tails can be arranged in 5!/(3!2!) = 10 ways, the probability of getting 3 heads and 2 tails is:

P(3 H’s and 2 T’s) = 10 x P(HHHTT) = 10 x 27/1024 = 5 x 27/512 = 135/512

Answer: B
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 16 Jun 2018, 00:47
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Probability of getting a head is 3 times the probability of getting a tail.

If probability of Head is X then Tails is (1-X)

x = 3(1-x) -> 3/4 = 3(1/4)

Probability of head = 3/4
probability of tail = 1/4

We are tossing 5 times for 3 heads:

5!/(3!*2!) * (1/4)^2 * (3/4)^3

= 10* (27/4^5) = (27*5)/4^4*2 = 135/512

OA-B.
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 07 Jul 2018, 01:02
I still don't understand this question. Can someone help?
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 07 Jul 2018, 01:23
Bunuel wrote:
prags1989 wrote:
I still don't understand this question. Can someone help?


If the probability that an unfair coin displays heads after it’s flipped is three times the probability that it will display tails, what is the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times?

A. 27/1024
B. 135/512
C. 9/64
D. 5/32
E. 11/32

A fair coin has the probability of heads equal to the probability of tails, so P(tail) = P(head) = 1/2. We are told that the coins is unfair so P(tail) ≠ P(head).

The probability that the coin displays heads after it’s flipped is three times the probability that it will display tails: P(h) = 3*P(t). Sin, the sum of these probabilities must be 1, then P(h) + P(t) = 1:

3*P(t) + P(t) = 1;

P(t) = 1/4 and P(h) = 3/4.

The question ask to find the probability that, after 5 coin flips, the coin will have displayed heads exactly 3 times. Exactly 3 heads can occur in several different ways:
HHHTT
HHTHT
HTHHT
THHHT
THHTH
THTHH
TTHHH
HHTTH
HTTHH
HTHTH
Basically, this is permutations of 5 letters HHHTT, out of which 3 H's and 2 T's are identical: 5!/(3!2!) = 10.

Now, each of the above 10 cases have the probability of (3/4)^3*(1/4)^2. Thus, the overall probability of P(HHHTT) is \(10*(\frac{3}{4})^3*(\frac{1}{4})^2 =\frac{135}{512}\).

Answer: B.

Hope it's clear.


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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 07 Jul 2018, 07:44
prags1989 wrote:
I still don't understand this question. Can someone help?

Basically this coins does not have equal chance of getting Head and Tails. Chance of getting Head is three times than getting Tail.
You have to consider different chances of getting Head and Tail.
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Re: If the probability that an unfair coin displays heads after it’s flipp  [#permalink]

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New post 05 Dec 2019, 23:25
Since we are dealing with an unfair coin, we understand that getting a head and a tail will not be equally likely, which is what the first statement of the question says as well.

Let the probability of getting a head = P(h) and the probability of getting a tail = P(t). Given that P(h) = 3* P(t).
The sum of the probabilities, P(h) + P(t) = 1.
Solving the two equations above, we get P(h) = ¾ and P(t) = ¼.

Now, in 5 coin flips, we need heads exactly three times. This means that the other two outcomes should be tails. So, the required combination of 3 heads and 2 tails is {H, H, H, T, T}.

There are two things that we need to note about the combination shown above:

1) The probability of this combination is \(\frac{3}{4}^3\) * \(\frac{1}{4}^2\) = \(\frac{27}{1024}\)

2) The heads and tails can be permuted in \(\frac{5!}{{3!*2!}}\) = 10 ways i.e. there are 10 different cases of obtaining 3 heads and 2 tails when the coin is flipped 5 times.

Therefore, the total probability = Probability of the combination * Number of permutations.
Substituting the respective values, required probability = \(\frac{27 }{ {1024}}\) * 10 = \(\frac{135 }{ 512}\).

The correct answer option is B.

Hope that helps!
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Re: If the probability that an unfair coin displays heads after it’s flipp   [#permalink] 05 Dec 2019, 23:25
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