What is the probability of rolling a number less than 3 at least 3 tim

Question

What is the probability of rolling a number less than 3 at least 3 times in 5 rolls of a six-sided die?

A. 2/15

B. 17/81

C. 1/3

D. 2/5

E. 15/32

IanStewart · Accepted Answer

This is an ugly question, and I think the best answer to it is "who cares?" These kinds of questions mislead test takers -- the GMAT is not testing if you're a human supercomputer who can work through three somewhat complicated cases in two minutes.

One thing to notice: there's a 1/3 chance on each roll that we get a small number, '1' or a '2', and a 2/3 chance we get a big number, from '3' through '6'. So the probability of any one specific sequence of five rolls will just be a product of five fractions that look either like "1/3" or "2/3". So the denominator of that product will be 3^5, and when we add up all our cases, we'll get a denominator of 3^5. We might be able to cancel out some of those 3's, but when you cancel down a fraction that starts out looking like k/3^5, you can never get some brand new prime in your denominator. So the answer to this question could never be 2/5 or 15/32 or 2/15. So only B and C are plausible answers, and since it doesn't seem all that likely we'll get the rare numbers (1 and 2) many times (at least 3 times) on five rolls, C seems too large, so B is the only answer that looks reasonable, but I'd still want to check, since that estimate isn't necessarily going to be reliable.

To compute exactly: 
• the probability all five rolls are small numbers is (1/3)^5
• the probability we get this sequence: large, small, small, small, small is (2/3)(1/3)^4, but we have 5 choices for when the large roll can happen, so we'll get exactly four small numbers with a probability of (5)(2/3)(1/3)^4
• the probability we get this sequence: large, large, small, small, small is (2/3)^2 (1/3)^3, but we have 5C2 = (5)(4)/2! = 10 choices for when the two large rolls can occur, so we'll get exactly three small numbers with a probability of (10)(2/3)^2(1/3)^3

and we finally get the answer by adding all three cases:

(1/3)^5 + (5)(2/3)(1/3)^4 + (10)(2/3)^2(1/3)^3 = (1 + 10 + 40)/3^5 = 51/3^5 = 17/3^4 = 17/81

CrackverbalGMAT · Answer

I was seeing if there was any better way of doing this sum other than by using the Binomial theorem. Guess that is the best approach.

Binomial Probability =  ^nC_x * (p)^x * (1-p)^{n-x} , where n = number of trials and x is the number of successes.

We want a number less than 3 i.e 1 or 2.  p = \frac{Favorable \space Outcomes}{Total \space Outcomes} = \frac{2}{6} = \frac{1}{3} \space and \space 1- p = \frac{2}{3}

Probability of getting less than 3 in at least 3 of the 5 rolls, which means we want a 1 or a 2 on 3 of the faces or on 4 of them or on all 5 of them.

P(3) =  ^5C_3 * (\frac{1}{3})^3 * (\frac{2}{3})^{5-3} = 10 * (\frac{1}{3})^3 * (\frac{2}{3})^2 = \frac{40}{243}

P(4) =  ^5C_4 * (\frac{1}{3})^4 * (\frac{2}{3})^{5-4} = 5 * (\frac{1}{3})^4 * (\frac{2}{3})^1 = \frac{10}{243}

P(5) =  ^5C_5 * (\frac{1}{3})^5 * (\frac{2}{3})^{5-5} = 1 * (\frac{1}{3})^5 * (\frac{2}{3})^0 = \frac{1}{243}

The required probability = P(3) + P(4) + P(5) =  \frac{40}{243} + \frac{10}{243} + \frac{1}{243} = \frac{51}{243} = \frac{17}{243}

Option B

Arun Kumar

SHoylandBPrep · Answer

Divide it into three cases:

1. Three rolls less than 3, Two rolls 3 or greater
2. Four rolls less than 3, One roll 3 or greater
3. Five rolls less than 3

1. Three rolls less than 3, Two rolls 3 or greater 
Calculate the probability of the rolls, LLLGG. The probability of L is  \frac{2}{6} = \frac{1}{3}  and the probability of G is  \frac{4}{6} = \frac{2}{3} .

→ Probability =  \frac{1}{3} × \frac{1}{3} × \frac{1}{3} × \frac{2}{3} × \frac{2}{3} = \frac{4}{243}

Now find the number of ways of rolling 3 Ls and 2 Gs. You are choosing 3 out of 5 dice to be L, so you can use the Combinations formula. (You can also think of it as rearranging the 'word' LLLGG.)

→ Combinations =  \frac{5!}{2!3!} = \frac{5 × 4}{2} = 10

Therefore, the probability is  \frac{40}{243} .

2. Four rolls less than 3, One roll 3 or greater 
Calculate the probability of LLLLG.

→ Probability =  \frac{1}{3} × \frac{1}{3} × \frac{1}{3} × \frac{1}{3} × \frac{2}{3} = \frac{2}{243}

Now find the number of ways of rolling 4 Ls and 1 Gs.

→ Combinations =  \frac{5!}{1!4!} = 5

Therefore, the probability is  \frac{10}{243} .

3. Four rolls less than 3, One roll 3 or greater 
Calculate the probability of LLLLL.

→ Probability =  \frac{1}{3} × \frac{1}{3} × \frac{1}{3} × \frac{1}{3} × \frac{1}{3} = \frac{1}{243}

There is only one way of rolling 5 Ls. Therefore, the probability is  \frac{1}{243} .

Answer 
Thus, the answer is

→ Probability =  \frac{40}{243} + \frac{10}{243} + \frac{1}{243} = \frac{51}{243} = \frac{17}{81}

Vivek1707 · Answer

A faster approach with less fractions

L stands for less than 3
G stands for greater than 3

Case 1 L L L GG ---> * 5!/(3!*2!) 
Case 2 L L L L G ----> * 5!/(4!)
Case 3 L L L L L ----> * 5! / 5!

Now note that for Less than 3 there are only two cases that the dice can take which is 1,2
For greater than 3 the dice can take 4 cases i.e 3 , 4 , 5 , 6

Case 1: 2^3 x 4^2 * 10 which is 2^7 x 10 
Case 2: 2^4 x 4^1 * 5 which is 2^6 x 5
Case 3: 2^5 * 1 which is 2^5

Denominator will have 6^5

So you get (2^5 x 10 + 2^6 x 5 + 2^5) / 2^5 x 3^5

2^5 gets factored out and cancelled

Left with (40 + 10 + 1) / 3^5 which is 17 /81

What is the probability of rolling a number less than 3 at least 3 tim

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