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# In tossing 4 fair dice, what is the probability of tossing,

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Manager
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In tossing 4 fair dice, what is the probability of tossing, [#permalink]

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27 Oct 2005, 18:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In tossing 4 fair dice, what is the probability of tossing, at most, one 3?
VP
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27 Oct 2005, 22:27
probability of no 3's = 5/6 * 5/6 * 5/6 * 5/6 = 625 / 1296
probability of one 3: 1/6 * 5/6 * 5/6 * 5/6 * 4 = 500 / 1296

together: 1125 / 1296
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28 Oct 2005, 14:26
To get "at most one 3" on 4 rolls of dice we have these 5 possibilities:
No 3: xxxx (x not= 3)
One 3: 3xxx, x3xx, xx3x, xxx3

p(xxxx) = (5/6)*(5/6)*(5/6)*(5/6) = 5^4/6^4
p(one 3) = (1/6)*(5/6)*(5/6)*(5/6) = 5^3/6^4

p(at most one 3) = 5^4/6^4 + 4*(5^3/6^4)

Simplify this further for the lack of A/C's..
= (5^4 + 4*5^3)/6^4
= (5*5^3 + 4*5^3)/6^4
= (5+4)*5^3/6^4
= 9*5^3//(3*2)^4
= 3^2*5^3/3^4*2^4
= 5^3/3^2*2^4
= 125/9*16
= 125/144
very high probability indeed
Manager
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28 Oct 2005, 14:29
BTW cloaked_vessel, you seem to have a rich collection of probability q's

cloaked_vessel wrote:
In tossing 4 fair dice, what is the probability of tossing, at most, one 3?
Manager
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28 Oct 2005, 18:22
nice work, oa is .868

Yes I am reviewing probability from a college text book. I am trying random problems in the back - some may not be direct gmat material but its practice nonetheless.
Director
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29 Oct 2005, 19:01
What would be the probability of getting atleast one 3. I am just trying to clear my concepts on alteast, atmost.. for probability.. its confusing..
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Success is my only option, failure is not -- Eminem

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29 Oct 2005, 19:19
bewakoof,
The probability of getting at least one 3 in three rolls is the same as:

1 - the probability of getting no 3's in 3 rolls...

Since there is a 1/6 probability that you will get at least one 3, the probability that you will get no 3s is 5/6. Thus, the prob that on all 3 rolls the die will not yield a 3 is:

5/6 * 5/6 * 5/6 = 125/216

Now we need to subtract this from one. The prob that at least one of the rolls will be a 3 includes every outcome, except when three consecutive non-3s are rolled. Therefore:

1- 125/216 = 91/216 = the prob that at least one 3 will be rolled

(I have illustrated the idea using three rolls) The question above involves four. I thought it was three and i don't want to go back and change it The same concept can be used though.
Hope this helps!
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31 Oct 2005, 08:54
cloaked_vessel wrote:
In tossing 4 fair dice, what is the probability of tossing, at most, one 3?

P(at most one 3) = P(no 3's) + P(one 3) * 1C4

P(no 3's) = 5/6*5/6*5/6*5/6=5^4/6^4

P(one 3) = 1/6*5/6*5/6*5/6*=5^3/6^4
---
P(at most one 3) = 5^4/6^4 + (4*5^3)/6^4 =

simplify by taking 5^3 out of brackets:

5^3(5 + 4) / 6^4 =

simplify by factoring denominator:

5^3 * 9 / 2^4 * 3^2 * 3^2 =

simplify by reducing by in in numerator and 3^2 in denominator:

5^3 / 2^4 * 3^2 = 125/16*9 = 125/144
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09 Nov 2005, 19:25
This can also be solved in the "opposite direction":

1 - P(of four 3's) - P(of three 3's) - P(of two 3's) = 1/6^4 - 20/6^4 - 25/6^3 = 1125/1296

Ugly, but correct.
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09 Nov 2005, 21:51
Find probability of throwing 0 threes + 1 threes

P( 0 threes) = (5/6)^4

P( 1 threes) = (1/6) * (5/6)^3

P (0 threes) + P( 1 threes) = (5/6)^3
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10 Nov 2005, 05:12
mbaqst wrote:
To get "at most one 3" on 4 rolls of dice we have these 5 possibilities:
No 3: xxxx (x not= 3)
One 3: 3xxx, x3xx, xx3x, xxx3

p(xxxx) = (5/6)*(5/6)*(5/6)*(5/6) = 5^4/6^4
p(one 3) = (1/6)*(5/6)*(5/6)*(5/6) = 5^3/6^4

p(at most one 3) = 5^4/6^4 + 4*(5^3/6^4)

Simplify this further for the lack of A/C's..
= (5^4 + 4*5^3)/6^4
= (5*5^3 + 4*5^3)/6^4
= (5+4)*5^3/6^4
= 9*5^3//(3*2)^4
= 3^2*5^3/3^4*2^4
= 5^3/3^2*2^4
= 125/9*16
= 125/144
very high probability indeed

I am getting 125/216. Add p(xxxx) and p(one 3)
125/216
10 Nov 2005, 05:12
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