What is the probability of rolling less than 3?

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In daily life, usually the word ‘probably’ is used when people are not sure about certain things. For example, Probably, India may win the match today. There may be a chance that India may win or lose or maybe the match is a tie. This type of statement leads to the uncertainty of the event. The word Probability is formed from the word word ‘probably’, which means that when people are not sure about an event is happens or not. People have a proper method to find out the probability which will be discussed in this article.

Terms used in Probability

Random Experiment: In the random experiment, we can not predict the result in advance. For example, if we toss a coin we can not predict that the head will appear, the tail also may appear.
Event: Collection of some outcome of an experiment is known as an event.
Sample Space: It is the collection of all possible outcomes. Suppose a dice is rolled out then the possible outcome is 1, 2, 3, 4, 5, or 6. It is denoted by S. S = (1, 2, 3, 4, 5, 6)
Rolling of Dice: A dice is a solid cube shape. It has 6 square faces. The six faces are marked by 1, 2, 3, 4, 5, 6 dots. When a fair dice is rolled, then the total possible outcome is 1, 2, 3, 4, 5, or 6. So all these numbers are known as sample space.

Probability

The probability of an event is defined as the ratio of favorable outcomes to the sample space or total outcomes. We represent it by ‘P’.
Probability of an event (P) = ( Number of Favourable outcomes) / (Total number possible outcomes)

Solution:

Concept: To solve the given problem, follow the steps given below.
Step 1: First of all find out all possible outcomes of the given event. Represent it by S.
Step 2: Specify the number of favorable outcomes.
Step 3: Use the formula, Probability of an event = (Favorable outcomes) / (Total number of possible outcomes)
Step 4: Simplify and get the final answer.
When a dice is rolled, all possible outcomes are 1, 2, 3, 4, 5, 6.
We call it as sample space, S = (1, 2, 3, 4, 5, 6)
So total number of possible outcomes = 6
Favorable outcome (Required outcome) = 1
(Only 1 is smaller than 2, remaining number is greater than 2 so we will not consider them as favorable outcomes.)
So total number of favorable outcomes = 1
Probability = (Total number of favorable outcomes)/(Total number of possible outcomes)
Probability = 1/6
So, the probability of the given statement is 1/6.

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What is the probability of rolling a number less than 3 at least 3 times in 5 rolls of a six-sided die?A. 2/15B. 17/81C. 1/3D. 2/5

E. 15/32

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Re: What is the probability of rolling a number less than 3 at least 3 tim [#permalink]

09 Oct 2020, 19:56

Bunuel wrote:

What is the probability of rolling a number less than 3 at least 3 times in 5 rolls of a six-sided die?A. 2/15B. 17/81C. 1/3D. 2/5

E. 15/32

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

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Re: What is the probability of rolling a number less than 3 at least 3 tim [#permalink]

10 Oct 2020, 07:51

Bunuel wrote:

What is the probability of rolling a number less than 3 at least 3 times in 5 rolls of a six-sided die?A. 2/15B. 17/81C. 1/3D. 2/5

E. 15/32

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)^3and 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 _________________

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What is the probability of rolling a number less than 3 at least 3 tim [#permalink]

10 Oct 2020, 09:58

Bunuel wrote:

What is the probability of rolling a number less than 3 at least 3 times in 5 rolls of a six-sided die?A. 2/15B. 17/81C. 1/3D. 2/5

E. 15/32

Divide it into three cases:1. Three rolls less than 3, Two rolls 3 or greater2. Four rolls less than 3, One roll 3 or greater3. 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 = \(\frac{40}{243} + \frac{10}{243} + \frac{1}{243} = \frac{51}{243} = \frac{17}{81}\)

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Re: What is the probability of rolling a number less than 3 at least 3 tim [#permalink]

20 Oct 2021, 00:44

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Re: What is the probability of rolling a number less than 3 at least 3 tim [#permalink]

20 Oct 2021, 00:44