It is currently 21 Oct 2017, 03:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 14 Dec 2004
Posts: 117

Kudos [?]: 67 [0], given: 0

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

### Show Tags

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?

Kudos [?]: 67 [0], given: 0

VP
Joined: 22 Aug 2005
Posts: 1111

Kudos [?]: 121 [0], given: 0

Location: CA

### Show Tags

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

Kudos [?]: 121 [0], given: 0

Manager
Joined: 17 Sep 2005
Posts: 72

Kudos [?]: 9 [0], given: 0

Location: California

### Show Tags

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

Kudos [?]: 9 [0], given: 0

Manager
Joined: 17 Sep 2005
Posts: 72

Kudos [?]: 9 [0], given: 0

Location: California

### Show Tags

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?

Kudos [?]: 9 [0], given: 0

Manager
Joined: 14 Dec 2004
Posts: 117

Kudos [?]: 67 [0], given: 0

### Show Tags

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.

Kudos [?]: 67 [0], given: 0

Director
Joined: 24 Oct 2005
Posts: 572

Kudos [?]: 76 [0], given: 0

Location: NYC

### Show Tags

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..
_________________

Success is my only option, failure is not -- Eminem

Kudos [?]: 76 [0], given: 0

Senior Manager
Joined: 05 Oct 2005
Posts: 485

Kudos [?]: 6 [0], given: 0

### Show Tags

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!

Kudos [?]: 6 [0], given: 0

Intern
Joined: 30 Oct 2005
Posts: 19

Kudos [?]: [0], given: 0

### Show Tags

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

Kudos [?]: [0], given: 0

Intern
Joined: 30 Oct 2005
Posts: 19

Kudos [?]: [0], given: 0

### Show Tags

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.

Kudos [?]: [0], given: 0

Senior Manager
Joined: 07 Jul 2005
Posts: 402

Kudos [?]: 61 [0], given: 0

### Show Tags

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

Kudos [?]: 61 [0], given: 0

Manager
Joined: 30 Aug 2005
Posts: 184

Kudos [?]: 15 [0], given: 0

### Show Tags

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

Kudos [?]: 15 [0], given: 0

10 Nov 2005, 05:12
Display posts from previous: Sort by