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# Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h

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Joined: 25 Mar 2011
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Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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Updated on: 09 Aug 2015, 00:16
2
13
00:00

Difficulty:

75% (hard)

Question Stats:

58% (02:38) correct 42% (02:42) wrong based on 102 sessions

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Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?

A. 20×(2^2/3^6)
B. 10×(2^3/3^6)
C. 20×(2^3/3^5)
D. 20×(2^3/3^6)
E. 20×(2^2/3^5)

Originally posted by smodak on 26 Jun 2011, 11:09.
Last edited by reto on 09 Aug 2015, 00:16, edited 1 time in total.
Added AC's and Source + proper format
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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26 Jun 2011, 11:43
1
2
Probability of getting 5 = 1/6
Probability of getting 6 = 1/6
Probability of getting 5 or 6 = P(5) + P(6)- P(5 and 6)
P(5 and 6) = 0 Since getting 5 and 6 is mutually exclusive event. i.e you cannot get both.
Probability of getting 5 or 6 = P(5) + P(6)
$$= 1/6 +1/6 = 2/6 = 1/3$$

Probability of getting (1,2,3,4) = 1-Probability of getting 5 or 6 = $$1-1/3 = 2/3$$

Probability of getting 5 or 6 exactly three times = $$(1/3) * (1/3) * (1/3) * (2/3) * (2/3) *(2/3)= 2^3/3^6$$

We need 5 or 6 to come exactly three times in sequence of 6 times dice roll.
(5 or 6) , (5 or 6), (5 or 6), (1,2,3,4), (1,2,3,4), (1,2,3,4)
we can get (5 or 6) in any order in 6 times for total of $$6C3= 20$$

Hence total ways = $$20 * 2^3/3^6$$
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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26 Jun 2011, 12:05
1
i have solved the example with normal principle of counting ...i.e. without using some formula or funda....

First of all the person tossed the dice 6 times hence the sample space would be 6^6

he got exactly 3 times either 5 or 6 i.e. 6C1+6C1 * 6C1+6C1 * 6C1+6C1 = 2*2*2

And in the remaining 3 tosses he got any number except 5 or 6 hence out of remaining 4 numbers he got any 1 = 4C1*4C1*4C1 = 4*4*4

hence in total of 6 tosses he got 2*2*2*4*4*4 = 2^9

Now, out of 6 tosses he got 5or6 in any 3 tosses thus it can be arranged in 6C3 ways = 20 ways

Now, Sample space is 6^6 , also, out of 6 tosses getting 5or6 is possible in 20 ways and results of the tosses are possible in 2^9 ways,

Thus chances of winning = 20 * 2^9 / 6^6 = 20* 2^9/ 2^6*3^6 = 20 * 2^3/ 3^6
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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26 Jun 2011, 20:47
2
smodak wrote:
Johnny the gambler tosses 6 plain dice. In order to win the jackpot he has to receive exactly 3 times a result of 5 or 6. What are Johnny's chances to win?

OA:
20×(2^3/3^6)

Source: Master GMAT

I use binomial theorem :
chances of 5 or 6 = 2/6 = 1/3
other numbers = 1-1/3= 2/3
Binomial theorem = nCr (p)^r*(1-p)^n-r
= 6C3(1/3)^3)(2/3)^6-3
= 20* 2^3/3^6
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Joined: 28 Mar 2011
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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26 Jun 2011, 22:02
sudhir18n wrote:
smodak wrote:
Johnny the gambler tosses 6 plain dice. In order to win the jackpot he has to receive exactly 3 times a result of 5 or 6. What are Johnny's chances to win?

OA:
20×(2^3/3^6)

Source: Master GMAT

I use binomial theorem :
chances of 5 or 6 = 2/6 = 1/3
other numbers = 1-1/3= 2/3
Binomial theorem = nCr (p)^r*(1-p)^n-r
= 6C3(1/3)^3)(2/3)^6-3
= 20* 2^3/3^6

A nice application of binomial theorem sudhir18n.
It did not strike me.

Thanks!

Regards,
Divya
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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09 Aug 2015, 00:19
1
smodak wrote:
Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?

A. 20×(2^2/3^6)
B. 10×(2^3/3^6)
C. 20×(2^3/3^5)
D. 20×(2^3/3^6)
E. 20×(2^2/3^5)

A nice question. I think it is worth of discussing again. Here I have the official answers:

This looks like a tough nugget, but let's break it down. First of all, look at one possible arrangement (h = 5 or 6, l = 1,2,3 or 4):

h,h,h,l,l,l (h=high, l=low)

The probability of getting 5 or 6 in a die toss is 1/6+1/6 = 1/3.
The probability of getting 1,2,3 or 4 in a die toss is 2/3.

So the odds of getting that particular arrangement are:

1/3*1/3*1/3*2/3*2/3*2/3 = 2^3/3^6

This is the chance for any given arrangement of three h and three l. However, this is only one possible scenario of having high results on three rolls. How many such scenarios exist?

What you need now is the number of possible arrangements: the number of ways of picking 3 high results out of 6 tosses. Order of choice doesn't matter, because you only care about which rolls are chosen - not the order you chose them in. For example, if you chose tosses 1,2,3 to yield high results, it does not matter in what order you chose them. Now that we have reduced the problem to a simple case of choosing k=3 out of n=6, order doesn't matter, use the Combinations formula:

6!/3!*3! = 20

So the probability that you're looking for is the number of arrangements times the probability of each arrangement.

I am happy to help.
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h  [#permalink]

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10 Jun 2018, 11:47
smodak wrote:
Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?

A. 20×(2^2/3^6)
B. 10×(2^3/3^6)
C. 20×(2^3/3^5)
D. 20×(2^3/3^6)
E. 20×(2^2/3^5)

Probability of getting a 5 or 6 on a single dice toss = 1/6 + 1/6 = 2/6

Now Probability of getting a 5 or 6 on exactly 3 of the 6 dice tossed = (2/6) * (2/6) * (2/6) * (1-2/6) * (1-2/6) * (1-2/6) * (6!/3!*3!) = 20*(2^3/3^6)

Thanks,
GyM
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Re: Johnny the gambler tosses 6 plain dice. In order to win the jackpot, h &nbs [#permalink] 10 Jun 2018, 11:47
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