Question

In two dice game, the player take turns to roll both dice, they can roll as many times as they want

in one turn. A player scores the sum of the two dice thrown and gradually reaches a higher score as

they continue to roll. If a single number 1 is thrown on either die, the score for that whole turn is

lost. Two dice are thrown simultaneously

37. What is the probability of getting the sum as an even number ?

(i) ¾ (ii) 1/2 (iii) 1/4 (iv) 5/8

38. What is the probability of getting the sum as a prime number ?

(i) 5/12 (ii) 1/6 (iii) 7/12 (iv) 11/12

39. What is the probability of getting the sum of at least 10?

(i) 5/12 (ii) 5/6 (iii) 1/6 (iv) 7/12

40. What is the probability of getting a doublet of even number ?

(i) 1/12 (ii) 5/12 (iii) 11/12 (iv) 7/12

41. What is the probability of getting a product of numbers greater than 16?

(i) 7/36 (ii) 2/9 (iii) 1/4 (iv) 11/36​

Answer 1

Step-by-step explanation:

Let's number all the throws of the die.

A can win only in odd throws, which are of the form (2i+1).

P(A wins) = ∑∞i=0P(A wins in the (2i+1)th throw)

because these events are mutually exclusive.

Note that the term corresponding to i=0 has value zero.

For other terms, we can write:

P(A wins in the (2i+1)th throw)

=P(a1+a2≠7)P(a2+a3≠7)...P(a2i−1+a2i≠7)P(a2i+a2i+1=7)

where aj is the number obtained in the jth throw.

Answer 2

Answer:

1. The probability of getting the sum as an even number = $\frac{1}{2}$

2. The probability of getting the sum as a prime number = $\frac{5}{12}$

3. The probability of getting the sum of at least 10 = $\frac{1}{6}$

4. The probability of getting a doublet of even number = = $\frac{1}{12}$

5. The probability of getting the product greater than 16 = $\frac{5}{18}$

Step-by-step explanation:

Recall the formula,

Probability of occurring an event = $\frac{No \ of\ favourable\ outcomes}{Total \ number\ of \ outcomes}$

When two dice are thrown simultaneously, then the possible outcomes 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)

Hence, Total number of outcomes = 36

1. The possible outcomes of getting the sum as an even number are

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

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

No. of possible outcomes of getting  the sum as an even number = 18

∴The probability of getting the sum as an even number = $\frac{18}{36} = \frac{1}{2}$

2. Probability of getting a sum as a prime number

= Probability of getting a sum equal to 2,3,5,7,11

Possible outcomes of getting a sum as a prime number are

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

No. of possible outcomes of getting  the sum as a prime number = 15

∴The probability of getting the sum as a prime number = $\frac{15}{36} = \frac{5}{12}$

3. Probability of getting the sum of at least 10 =  Probability of getting a sum equal to 10,11,12

Possible outcomes of getting a sum equal to 10,11,12 are

(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)

No. of possible outcomes of getting  the sum of at least 10 = 6

∴The probability of getting the sum of at least 10 = $\frac{6}{36} = \frac{1}{6}$

4.  Possible outcomes of getting a doublet of even number =

(2,2),(4,4),(6,6)

No. of possible outcomes of getting a doublet of even number = 3

∴The probability of getting a doublet of even number = $\frac{3}{36} = \frac{1}{12}$

5.  Possible outcomes of getting the product of numbers greater than 16 =(3,6),(4,5),(4,6), (5,4), (5,5),(5,6), (6,3), (6,4), (6,5), (6,6)

No. of possible outcomes of getting the product of numbers greater than 16 = 10

∴The probability of getting the product of numbers greater than 16 = $\frac{10}{36} = \frac{5}{18}$

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