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A coin has two sides. One side has the number 1 on it and the other

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A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
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[Reveal] Spoiler: OA

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
Kudos for a correct solution.


Cases for sum to be greater than 4 are as mentioned below

Case 1: (1, 2, 2)
Case 2: (2, 1, 2)
Case 3: (2, 2, 1)
Case 4: (2, 2, 2)

Total Outcomes = 2 x 2 x 2 = 8

Probability = Favorable Outcomes / Total Outcomes

I.e. Probability = 4/8 = 1/2

Answer: Option
[Reveal] Spoiler:
D

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 25 Jun 2015, 15:20
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Drawing the probability tree, first throw 1 or 2 will come, 2nd throw 1 or 2 will come, 3rd throw 1 or 2 will come. Therefore there are only 4 ways out of 8 for which sum of numbers will be greater than 4.
Cases:
1-2-2
2-1-2
2-2-1
2-2-2
Hence probability = 4/8 =1/2

Hence answer is D
Thanks,

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 25 Jun 2015, 18:15
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Hi All,

The number of total outcomes and outcomes that give us what we "want" are relatively low, so calculating the answer to the given question isn't too difficult or time-consuming. In these types of prompts, it's sometimes faster to calculate what we "DON'T want" and then subtract that from 1 (to determine the probability of what we DO want).

Here, there are two outcomes that will NOT match what we're looking for: a sum or 3 or a sum of 4....

Ways to get a sum of 3:
1-1-1

Ways to get a sum of 4:
1-1-2
1-2-1
2-1-1

Each coin has two possible outcomes, so there are (2)(2)(2) = 8 possible permutations.

4/8 = probability of what we DON'T want

1 - 4/8 = probability of what we DO want

1 - 4/8 = 4/8 = 1/2

Final Answer:
[Reveal] Spoiler:
D


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A coin has two sides. One side has the number 1 on it and the other [#permalink]

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Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
Kudos for a correct solution.



total ways =\(2^3\)

ways to get >4..
1)2,2,2 - 1 way
2) 2,2,1 -3!/2!=3 ways
total -4 ways

prob=4/8=1/2
ans D
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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
Kudos for a correct solution.


My attempt -
There are 2 possible outcome in each toss and coin is tossed for 3 times,
Hence total Outcome = \(2^3\)

Now Result of 3 toss will be greater than 4, only if we have at-least 2 for two times out of 3 times.

That means,
Required Outcome = {At least 2 twos) + {All twos) = Total combination of {At least 2 twos) + Total combination of {All twos}

Total combination of {At least 2 twos) = 3! / 2! * 1! = 3 ----- (Since two 2s are identical hence 2! in denominator)
Total combination of {All twos} = 3! / 3! = 1

Hence,
Required Probability = \frac{(Required Outcome)}{Total outcome} = \frac{(3+1)}{8} = \frac{1}{2}

Option D

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 29 Jun 2015, 05:00
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Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
Kudos for a correct solution.


Platinum GMAT Official Solution:

One approach to solve the problem is to list the different possibilities for a toss of coin three times. Because there are two outcomes and the coin is tossed three times, the table will have 2*2*2 or 8 rows.

Next add the resulting rows together to find the sum (the fourth column in the table below).

Toss 1 | Toss 2 | Toss 3 | Sum
1 ---------- 1 -------- 1 ------ 3
1 ---------- 1 -------- 2 ------ 4
1 ---------- 2 -------- 1 ------ 4
1 ---------- 2 -------- 2 ------ 5
2 ---------- 1 -------- 1 ------ 4
2 ---------- 1 -------- 2 ------ 5
2 ---------- 2 -------- 1 ------ 5
2 ---------- 2 -------- 2 ------ 6

From the table we see that there are 4 situations where the sum of the tosses will be greater than 4. And there are 8 possible combinations resulting in a probability of

4/8 or a probability of 1/2.

SO the correct answer is D.
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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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Total=2*2*2=8
Favorable = 4 ( 2,2,2 2,2,1 2,1,2 1,2,2)
Probability=4/8 =1/2
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A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 22 Oct 2016, 21:51
Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4

Source: Platinum GMAT
Kudos for a correct solution.


2 sides of coin flipped 3 times \(2^3\) = 8 outcomes
111,112,121,211 whose sum \(<=\) 4
greater than 4
favorable 8-4=4
\(\frac{4}{8}\) or \(\frac{1}{2}\)
Answer is D

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 29 Aug 2017, 04:00
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4? --> I have an issue with the wording and solution hence: doesn't it say more than 4? Which means sum of 5 and above?
But the solution includes 4 as a sum too
:(
In such a case, where question says greater than 'certain number' we include it unless specified otherwise?

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Re: A coin has two sides. One side has the number 1 on it and the other [#permalink]

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New post 01 Sep 2017, 12:06
Bunuel wrote:
A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

A) 3/8
B) 1/16
C) 1/8
D) 1/2
E) 1/4


We can use the equation:

1 = p(sum greater than 4) + p(sum not greater than 4)

The only way to get a sum that is not greater than 4 is to get 1-1-1 or 2-1-1.

P(1-1-1) = 1/2 x 1/2 x 1/2 = 1/8.

P(2-1-1) = 1/2 x 1/2 x 1/2 = 1/8.

Since there are 3 ways to flip a 2, 1, and 1 (they are 2-1-1, 1-2-1, 1-1-2), the probability P(2-1-1) is actually 3/8.

Thus:

P(sum greater than 4) = 1 - (3/8 + 1/8) = 1 - 4/8 = 1/2.

Alternate solution:

The only way to get a sum that is greater than 4 is to get 2-2-2 or 2-2-1.

P(2-2-2) = 1/2 x 1/2 x 1/2 = 1/8.

P(2-2-1) = 1/2 x 1/2 x 1/2 = 1/8.

Since there are 3 ways to flip a 2, 2, and 1 (they are 2-2-1, 2-1-2, 1-2-2), the probability P(2-2-1) is actually 3/8.

Thus:

P(sum greater than 4) = 1/8 + 3/8 = 4/8 = 1/2.

Answer: D
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Re: A coin has two sides. One side has the number 1 on it and the other   [#permalink] 01 Sep 2017, 12:06
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