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

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Math Expert
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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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25 Jun 2015, 04:50
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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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25 Jun 2015, 05:25
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Expert's post
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

[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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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

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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25 Jun 2015, 18:15
1
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Expert's post
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

[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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25 Jun 2015, 20:28
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Expert's post
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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25 Jun 2015, 22:01
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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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29 Jun 2015, 05:00
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Expert's post
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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09 Nov 2015, 22:02
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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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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 $$\frac{1}{2}$$

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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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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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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.

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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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