Probability need help

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Winner2013: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Sun Jul 07, 2013 5:35 am

Probability need help

by Winner2013 » Sun Aug 25, 2013 11:19 am

On a biased dice,any even number appears four times as frequently as any odd number. If the dice is rolled thrice what is the probability that the sum of the scores on those is more than 16?

1. 26/375 2. 112/375 3. 26/3375 4. 112/3375 5. 8/225. Answer is option 4

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Source: Beat The GMAT — Problem Solving | Sun Aug 25, 2013 11:19 am

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by [email protected] » Sun Aug 25, 2013 12:30 pm

Hi Winner2013,

This question layers on the details, so you have to be very careful with how you organize it. There are a couple of different ways to do the math, but here's a way that allows you to see all of the possibilities:

First, let's deal with the fact that even numbers appear 4x as frequently as odd numbers.

Roll
1 x 1
2 x 4
3 x 1
4 x 4
5 x 1
6 x 4

Total possible outcomes per roll = 15

Now, we need to calculate the probability of what we WANT (a three-roll sum > 16): So, we need a total of 17 or 18.

To get a 17 or 18, we need one of the following possibilities:
665
656
566
666

Odds of 665: (4/15)(4/15)(1/15) = 4^2/15^3
Odds of 656: Same as above = 4^2/15^3
Odds of 566: Same as above = 4^2/15^3

Odds of 666: (4/15)(4/15)(4/15) = 4^3/15^3

Total odds: 16(3) + 64 = 112/15^3

Final Answer: 4

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

Contact Rich at [email protected]

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Winner2013: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Sun Jul 07, 2013 5:35 am

by Winner2013 » Sun Aug 25, 2013 1:05 pm

thank you Rich. I understood the logic better now. I have been following your posts and you help clarify concepts. thank you again. I would like to tell you what i thought about this problem just to understand how my thought process has been wrong, so that I don't repeat the same mistake again.

Chances of getting even are 4 times getting an odd.
So probability of getting even = 4/5
And p(odd) = 1/5

Now to get sum>16 it should be either 17 or 18
For 17- 3!/2! = 3 ways considering 3 arrangements
And 18 there is 1

So for 17 probability = (4/5 *4/5*1/5 ) *3
And for 18 = (4/5*4/5*4/5)

So total = (16*3 +64)/125 = 112/125

i don't know where am i going wrong. Please help. Appreciate it. Thank you

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by [email protected] » Sun Aug 25, 2013 3:30 pm

Hi Winner2013,

The reason why your logic didn't quite work in this question is that you did not account for ALL of the numbers. To score 17 or 18 on three rolls, you CAN'T have any 1s, 2s, 3s or 4s. Your math doesn't account for that; it only accounts for the probability of rolling an odd (1/5) vs. rolling an even (4/5).

You DID correctly deduce that there are 3 ways to roll 17 and 1 way to roll 18. To adjust your calculation, you have to consider ALL the possible outcomes (so you divide each value by 15, not 5).

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

Contact Rich at [email protected]

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Winner2013: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Sun Jul 07, 2013 5:35 am

by Winner2013 » Sun Aug 25, 2013 5:18 pm

Yes you are right. Thank you very much Rich.

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ganeshrkamath: Master | Next Rank: 500 Posts; Posts: 283; Joined: Sun Jun 23, 2013 11:56 pm; Location: Bangalore, India; Thanked: 97 times; Followed by:26 members; GMAT Score:750

by ganeshrkamath » Sun Aug 25, 2013 9:26 pm

Winner2013 wrote:On a biased dice,any even number appears four times as frequently as any odd number. If the dice is rolled thrice what is the probability that the sum of the scores on those is more than 16?

1. 26/375 2. 112/375 3. 26/3375 4. 112/3375 5. 8/225. Answer is option 4

Probability of even number appearing = 4p
Probability of odd number appearing = p
4p*3 + p*3 = 1
p = 1/15
4p = 4/15

To get more than 16 (i.e. 17 or 18) the numbers should be (6,6,5) or (6,6,6) in any order.
P(6,6,6) = 4/15*4/15*4/15 = 64/3375
P(6,6,5) = 4/15*4/15*1/15 = 16/3375 = P(6,5,6) = P(5,6,6)
Total probability = 64/3375 + 3*16/3375 = (64+48)/3375
= 112/3375

Choose [spoiler](4)[/spoiler]

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