A die is rolled in such a way that an even number is twice as likely to occur as an odd number if the die is tossed twice find the probability distribution of random variable representing the number as prime number on the two tosses.
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Given : A die is rolled in such a way that an even number is twice as likely to occur as an odd number if the die is tossed twice
To find : Probability of number as prime number on the two tosses.
Solution:
A die is rolled in such a way that an even number is twice as likely to occur as an odd number
probability of odd number 1 , 3 ,5 = p
Probability of even number 2 , 4 , 6 = 2p
p + p + p + 2p + 2p + 2p = 1
=> p = 1/9
Probability of odd number = 1/9
Probability of even number = 2/9
possible numbers
11 , 12 , 13 , 14 , 15 , 16 , 21 , 22 , 23 , 24 , 25 , 26 , 31 , 32 , 33 , 34 , 35 , 36
41 , 42 , 43 , 44 , 45 , 46 , 51 , 52 , 53 , 54 , 55 , 56 , 61 , 62 , 63 , 64 , 65 , 66
Prime numbers - 11 , 13 , 23 , 31 , 41 , 43 , 53 , 61
(1/9)(1/9) + (1/9)(1/9) + (2/9)(1/9) + (1/9)(1/9) + (2/9)(1/9) + (2/9)(1/9) + (1/9)(1/9) + (2/9)(1/9)
= (1 + 1 + 2 + 1 + 2 + 2 + 1 + 2)/81
= 12/81
12/81 is the Probability
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