# dice is marked as 6 distinct odd numbers 1,3,5,7,9,11 such that sum of the opposite pairs is 12. if 10 similar dice are thrown together what is the possibility of having a sum as 55

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get dice is marked as 6 distinct odd numbers 1,3,5,7,9,11 such that sum of the opposite pairs is 12. if 10 similar dice are thrown together what is the possibility of having a sum as 55 from screen.

Question

Two dice are rolled.

If both dices have six faces numbered 1,2,3,5,7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is

Open in App Solution

The correct option is **A**

1736

**Explanation for the correct answer:**

Two dice are rolled.

The total number of outcomes in sample space nS are given as

⇒ nS=62=36

The numbers on the faces of the dice are 1,2,3,5,7,11

The possible scenarios where the sum of the digits is less than or equal to eight is with the combination of 1,2,1,3,1,5,1,7,2,3,2,5,3,5

Hence the ordered pairs with the combination of these digits are 1,1,1,2,1,3,1,5,1,7,2,1,2,2,2,3,2,5,3,1,3,2,3,3,3,5,5,15,25,3,7,1

Number of ordered pairs =17

Hence the requiredprobability=NumberoforderedpairsnS

requiredprobability=1736

Hence, the probability of the sum of the digits appearing on the faces of the dice less than or equal to eight is 1736.

**Hence, option**A

**is the correct answer.**

Conditional Probability

Standard XII Mathematics

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## Six faces of an unbiased die are numbered with 2, 3, 5, 7, 11 and 13 . If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is

Click here👆to get an answer to your question ✍️ Six faces of an unbiased die are numbered with 2, 3, 5, 7, 11 and 13 . If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is

Question

## Six faces of an unbiased die are numbered with 2,3,5,7,11 and 13. If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is

**A**

18 5

**B**

36 5

**C**

18 13

**D**

36 25 Medium Open in App Solution Verified by Toppr

Correct option is A)

The sum on the uppermost faces of the dice is an odd number. Hence exactly one of the two numbers on the uppermost face is 2. So the different possible combinations are (2,3),(2,5),(2,7),(2,11),(2,13),(3,2),(5,2),(7,2),(11,2) and (13,2)Total required combinations =10

The probability that the sum on the uppermost faces of the dice is an odd number is =

36 10 = 18 5

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16 3

## The faces of a die bear the numbers 1,3,5,7,9,11. The die is rolled. Find the probability of getting a perfect square number on the upper face of the dice. (i) S={1,3,5,7,9,11}" "(ii)thereforen(S)=6 (iii) Let A be the event of getting a perfect square number. Then P(A)=(n(A))/(n(S))=

(iii) Let A be the event of getting a perfect square number. Then A={1,9}therefore n(A)=2 (iv) P(A)=(n(A))/(n(S))=2/6=1/3.

Home > English > Class 10 > Maths > Chapter > Probability >

The faces of a die bear the nu...

The faces of a die bear the numbers

1,3,5,7,9,11. 1,3,5,7,9,11.

The die is rolled. Find the probability of getting a perfect square number on the upper face of the dice.

(i)S={1,3,5,7,9,11} (ii)∴n(S)=6

(i)S={1,3,5,7,9,11} (ii)∴n(S)=6

(iii) Let A be the event of getting a perfect square number. Then

P(A)= n(A) n(S) =−=− P(A)=n(A)n(S)=-=-

Updated On: 27-06-2022

00 : 30

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Text Solution Open Answer in App Solution

(iii) Let A be the event of getting a perfect square number. Then

A={1,9}∴n(A)=2 A={1,9}∴n(A)=2 (iv) P(A)= n(A) n(S) = 2 6 = 1 3 .

P(A)=n(A)n(S)=26=13.

Answer

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Guys, does anyone know the answer?