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If a coin has an equal probability of landing heads up or

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Richard Lee
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If a coin has an equal probability of landing heads up or [#permalink] New post   27 Dec 2007, 02:43
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If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?

A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
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C
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[#permalink] New post   28 Dec 2007, 19:39
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this also works....

we know that we want 2 H and 1 T, so we have 3 possibilities

HHT
HTH
THH

find the probability of each and then add....
HHT= 1/2*1/2*1/2 = 1/8
HTH= 1/2*1/2*1/2 = 1/8
THH= 1/2*1/2*1/2 = 1/8

1/8+1/8+1/8= 3/8 or 0.375
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[#permalink] New post   27 Dec 2007, 04:12
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C

P=nCm*p^m*q^(n-m)
n - the total number of trials.
m - the number of trials with heads.
n-m - the number of trials with tails.
p - probability of heads.
q - probability of tails.

P=3C2*(1/2)^2*(1/2)^1=3/8=0.375
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[#permalink] New post   29 Dec 2007, 07:40
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probability of 2 heads =3C2 *1/2^3 =3/8 =0.375
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Re: coin probability [#permalink] New post   27 Sep 2009, 11:34
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If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?

A 0.125
B 0.25
C 0.375
D 0.5
E 0.666

Soln:
Total number of ways in which H or T can appear in 3 tosses of coin is
= 2 * 2 * 2 = 8 ways

For 2 H and 1 T
HHT, HTH, THH

Thus probability is
= P(HHT) + P(HTH) + P(THH)
= 1/8 + 1/8 + 1/8
= 3/8
= .375

Ans is C
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Re: coin probability [#permalink] New post   29 Mar 2011, 08:56
It actually helps reading and understanding the problem correctly (:
I understood the question as 'what is the probability that the coin will land heads up consequently twice in 3 consecutive flips'?

In that case:
HHT = 1/8
THH = 1/8
Total outcomes: 8
Probability: 1/4
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Re: coin probability [#permalink] New post   29 Mar 2011, 19:23
3C2 * (1/2)^2(1/2)

= 3 * 1/8

= 0.375
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   20 Sep 2013, 03:28
You can either determine that there are three outcomes (HTT, THT, TTH) that fit the criteria of use the formula 3C2 (3!/2!(3!-2!)

each outcome has a probality of 1/2*1/2*1/2 = 1/8

3 outocomes with P=1/8 = 3/8

1/8=0.125 or 3/8 is 0.375
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If a coin has an equal probability of landing heads up or [#permalink] New post   19 Jan 2016, 13:42
Richard Lee wrote:
If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?

A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666


Probability * # of arrangements --> \(1/2^3*3=0,375\)
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If a coin has an equal probability of landing heads up or [#permalink] New post   Updated on: 13 Feb 2021, 08:39
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Richard Lee wrote:
If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?

A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666


We need to determine the probability of HHT and its variations:

P(HHT) = (1/2)^3 = 1/8

Since we can arrange HHT in 3!/2! = 3 ways, the overall probability is 3 x 1/8 = 3/8 = 0.375.

Answer: C

Originally posted by ScottTargetTestPrep on 22 Feb 2019, 16:37.
Last edited by ScottTargetTestPrep on 13 Feb 2021, 08:39, edited 1 time in total.
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   22 Feb 2019, 18:02
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Another way to do this: if you flip a coin an odd number of times, half the time you get more heads than tails, and half the time you get more tails than heads. You get three heads (1/2)^3 = 1/8 of the time, and since the only other way to get more heads than tails is to get two heads, that must happen 3/8 of the time.
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   04 Sep 2020, 08:16
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\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)


When a coin is flipped n times, the total number of outcomes = \(2^n\)

When it is flipped 3 times, the total outcomes = \(2^3 = 8\)

Favorable outcomes is when 2 out of the 3 flips lands on heads = 3C2 = 3 ways (HHT, HTH, THH)

The required probability = \(\frac{3}{8} = 0.375\)


Option C

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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   12 Oct 2020, 07:30
Probability of getting heads: 1/2
Probability of getting tails: 1/2

P(heads) * P(heads) * P(tails) =

1/2 * 1/2 * 1/2 = 1/8

The three events are mutually exclusive; thus, order does not matter here. There number of ways we can choose 2 coins from 3 is 3 ways.

1/8 * 3 = 3/8 = 0.375
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   09 Dec 2020, 05:30
Dear Expert,

What is the significance of the consecutive flips here?
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   13 Feb 2021, 01:01
Probability (Complex Event) = Probability (Simple event) * Number of Arrangements

Probability (Exactly 2 heads when coin tossed thrice) = ½ * ½ * ½ * (3!/2!) = 3/8 { THH, HTH and HHT}

So, the answer is C.
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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   13 Feb 2021, 16:25
P(H) = 1/2, P(T) = 1/2

Possible sequences:
HHT
HTH
THH

P(2H and 1T) = 1/2 x 1/2 x 1/2 = 1/8

1/8 + 1/8 + 1/8 <--- We add because there are three (mutually exclusive orders)

3/8

Answer is C.
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If a coin has an equal probability of landing heads up or [#permalink] New post   16 Oct 2022, 11:23
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We need to find On three consecutive flips of a coin, what is the probability that the coin will land heads up exactly twice in 3 consecutive flips

Coin is tossed 3 times => Total number of cases = \(2^3\) = 8

Lets solve the problem using two methods

Method 1:

Out of the 8 cases there are only three cases in which we get exactly two heads. HHT, HTH and THH.

=> Probability of getting exactly two heads = \(\frac{3}{8}\) = 0.375

So, Answer will be C

Method 2:

We have three places _ _ _ and we need to find the two places in which we can get a Head. We can do that in 3C2 ways
=> \(\frac{3!}{2! * 1!}\) = 3 ways

=> Probability of getting exactly two heads = Number of ways * P(H) * P(H) * P(T) = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\) = 0.375

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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Re: If a coin has an equal probability of landing heads up or [#permalink] New post   26 Dec 2023, 01:38
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