Question

Probability chapter

Answer the ques in the attachment with proper explanation


Class 10

Answer 1

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Hello Board Rocker !

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Hope you will find my Solution in the list of Brainlest !

The Combination of two dice is equal to 36 When We throw two dices

Total Outcomes  = 36

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( i) Favorable Outcomes are listed as below :

(2,2) (2,3) (2,5) (3,2) (3,3) (3,5) (5,2) (5,3) (5,5)  

Total 9 Favourable  Outcomes

Probability of a Prime on each dice

Probability = $\frac{Favorable \ Outcome }{Total \ Outcome }$

P = $\frac{9}{36}$

P = $\frac{1}{4}$

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(ii) A total of 9 and 11

Favorable Outcomes = ( 3,6) (4,5) (5,4) (6,3) (5,6) (6,5)

Total Favorable Outcome = 6

$P = \frac{6}{36}$

P = $\frac{1}{6}$

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(iii) A number Greater than 3 on each dice

P = $\frac{9}{36}$

P = $\frac{1}{4}$

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Note : I am Posting half part of the answer rest part (iv) , (v) and (vi) will Continue by Sid Bhaiya

Thanks for Reading !

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@FuturePoet

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Answer 2

Answer:

Step-by-step explanation:

Continuation to Futurepoet:

(iv)

Total number of possible outcomes n(S) = 6²

                                                                  = 36.

Let A be the event of getting a total of 6 or 7 on the number of two dice.

n(A) = {1,5},{2,4},{3,3},{4,2}, {5,1},{1,6},{2,5},{3,4},{4,3},{5,2},{6,1}

      = 11.

Therefore, Required probability P(A) = n(A)/n(S)

= 11/36.

(v)

Let B be the event of getting a Sum which is a prime number.

n(B) = {1,1},{1,2},{1,4},{1,6},{2,1},{2,3},{2,5},{3,2},{3,4},{4,1},{4,3},{5,2},{5,6},{6,1},{6,5}

      = 15

P(B) = 15/36

Let B' be the event of getting a sum which is not a prime number.

P(B') = 1 - P(B)

       = 21/36

       = 7/12.

(vi)

Not possible.

Hope it helps!