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

Question

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.

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

Source: Platinum GMAT 
 Kudos for a correct  solution .

GMATinsight · Answer

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  D

lipsi18 · Answer

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,

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made,
Rich

chetan2u · Answer

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

vishwaprakash · Answer

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

Bunuel · Answer

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.

jasimuddin · Answer

Total=2*2*2=8
Favorable = 4 ( 2,2,2 2,2,1 2,1,2 1,2,2)
Probability=4/8 =1/2

raven09 · Answer

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

Madhavi1990 · Answer

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.

Show Answer

Official Answer

D

In such a case, where question says greater than 'certain number' we include it unless specified otherwise?

JeffTargetTestPrep · Answer

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

NightWalker · Answer

A quicker approach but may not work in a more complex situation:
In order to get a sum of more than 4 out of 3 flips (assuming the coin is not biased, I.e. there is equal probability of landing 1 or 2): 
_ _ _ - 3 flips
The first 2 flips has to be 2. Let’s fix: 2,2,_
Since required sum of events should be more than 4, the last flip can either land as 1 or 2 so that the sum is more than 4.
And since the coin is not biased, it’ll have 1/2 probability of landing either 1 or 2 (in both cases, the sum will be more than 4).
Please let me know if you think my approach is crazy..🥴

Posted from my mobile device

EMPOWERgmatRichC · Answer

Hi Ashmin,

There are a couple of problems with your logic. First, if the sum of the three tosses has to be greater than 4, then the first two tosses do NOT necessarily have to be 2 and 2 (for example, if you had 1 and 2 and then got a 2 on the third toss, then you would have a total of 5). Second, if you are going to 'fix' the first two tosses as 2 and 2, then you still have to account for the probability of that event occurring. Since each side of the coin has the same chance of occurring, the probability of flipping 2 and 2 on the first two tosses is...

(1/2)(1/2) = 1/4

...meaning that there's only a 1 in 4 chance of the first two tosses both being 2s - and your approach doesn't account for that.
 
GMAT assassins aren't born, they're made,
Rich

BrushMyQuant · Answer

Given that 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. 
We need to find what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?

As the coin is tossed 3 times => Number of cases =  2^3  = 8 cases

Sum will be more than 4 when we get at least two 2's out of 3 tosses.
=> Following are the cases
(1, 2, 2), (2, 1, 2), (2, 2, 1), (2, 2, 2)
=> 4 cases

=>  Probability that the sum of the numbers on the landing side of the coin will be greater than 4  =  \frac{4}{8}  =  \frac{1}{2}

So,  Answer will be D 
Hope it helps!

The Concept is similar to that of Rolling a dice. Watch the following video to learn How to Solve Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

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