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in a simultaneous throw of a pair of dice, find the probability of getting a total more than 11?

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Question

In a simultaneous throw of a pair of dice, find the probability of getting a total more than 10.

A

12 1 ​

B

2 1 ​

C

9 1 ​

D

6 1 ​ Medium Open in App Solution Verified by Toppr

Correct option is A)

Two dice were thrown.

Total outcomes: 36

Number of outcomes which makes sum of more than 10 are: 3→{(5,6),(6,5),(6,6)}

Required probability:

P= 36 3 ​ = 12 1 ​

0 1

स्रोत : www.toppr.com

[Solved] In a simultaneous throw of a pair of dice, find the probabil

Calculation: When 2 dice are throw total outcome = 36 Number of outcomes where sum of throws is 9 = (4, 5), (5, 4), (3, 6) and (6, 3) = 4 Number of outcomes

Home Quantitative Aptitude Probability

In a simultaneous throw of a pair of dice, find the probability of getting a total of 9 or more.

5/18 5/12 2/7 2/36

Option 1 : 5/18

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Detailed Solution

Calculation:

When 2 dice are throw total outcome = 36

Number of outcomes where sum of throws is 9 = (4, 5), (5, 4), (3, 6) and (6, 3) = 4

Number of outcomes where sum of throws is 10 = (4, 6), (6, 4), and (5, 5) = 3

Number of outcomes where sum of throws is 11 = (6, 5), and (5, 6) = 2

Number of outcomes where sum of throws is 12 = (6, 6) = 1

Total number of ways of getting sum 9 and above = 4 + 3 + 2 + 1 = 10 ways

Probability of getting  sum 9 and above = 10/36 = 5/18

∴ Required probability is 5/18.

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स्रोत : testbook.com

In a simultaneous throw of two dice, what is The probability of getting a total of 10 or 11?

Answer (1 of 5): Sample space S- two dice thrown N(S) = 36 Event A- the total is 10 N(A)= 3 The probability of having 10 in total of the dice faces P(A)=N(A)/N(S) P(A)=1/12 Similarly for having 11 on face we get probability 1/18

In a simultaneous throw of two dice, what is The probability of getting a total of 10 or 11?

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Total number of possible outcomes, N = 36

Let A = The sum is 10 or 11 = { (4, 6), (4, 6), (5, 5), (5, 6), (6, 5) }

N(A) = 5

Therefore, P(A) = 5/36

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There are a total of 36 combinations in throw of 2 dice.

The sum of 10 or 11 could be obtained in the following cases:

5 + 5, 4 + 6, 6 + 4 5 + 6, 6 + 5

That is a total of 5 cases. Therefore the required probability here is:

= 5/ 36 Sponsored by Aspose

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High volume IBM wafer fab experience in process, metrology and defect areasAuthor has 128 answers and 474.5K answer views5y

The probability of getting a total of 10 or 11 with two die thrown simultaneously is %13.89.

This breaks down as:

Getting a total of 10 is possible with a 4+6, 5+5 or a 6+4 for 3 of 36 possible combinations.

Getting a total of 11 is possible with a 5+6 or a 6+5 for 2 of 36 possible combinations.

Total probability is 5/36 or %13.89

Here is a link to show you how this works:

Probability - throwing dice

Nigel Parsons

Studied at Bishop of Llandaff Church in Wales High SchoolAuthor has 1.3K answers and 663.6K answer views1y

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In a simultaneous throw of two dice, what is the probability of getting a total of 8 or 9?

In a simultaneous throw of two dice, what is the probability of getting a total of 8 or 9?

In a simultaneous throw of two fair dice (each numbered 1–6) There are 36 (6*6) possible combinations.

There are 5 ways of scoring 8 (2,6 3,5 4,4 5,3 6,2) so the probability of scoring 8 is 5/36

There are 4 ways of scoring 9 (3,6 4,5 5,4 6,3) so the probability of scoring 9 is 4/36 = 1/9

The probability of getting a total of either 8 or 9 is the sum of these two probabilities: 5/36 + 4/36 = 5+4/36 = 9/36 = 1/4 or 25%

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Ronnie Wayne

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5/36

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What is the probability of rolling a sum of 7 with two dice?

When 2 dice are rolled total possible outcomes are 36. They are :-

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Total favourable outcomes to get a sum of 7 when 2 dice are rolled simultaneously = 6

{i.e.,(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Probability = favourable outcomes /total outcomes

P = 6/36 P = 1/6.

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Saurabh Salekar

Bsc in Bsc (hons) Statistics, Elphinstone College, Fort, Mumbai5y

Sample space S- two dice thrown

N(S) = 36

Event A- the total is 10

N(A)= 3

The probability of having 10 in total of the dice faces

P(A)=N(A)/N(S) P(A)=1/12

Similarly for having 11 on face we get probability 1/18

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