Two unbiased dice are thrown. find the probability that (i) the first die shows 6. (ii) the total of the numbers on the dice is 8. (iii) both the dice show the same number. (iv) the total of the numbers on the dice is greater than 8. (v) the total of the numbers on the dice is 13. (vi) the total of the numbers on the dice is any number from 2 and 12, both inclusive
Answers
(ii)=P(E)=5/36
(iii)=P(E)=0
(iv)=P(E)=1
All Probabilities found
Step-by-step explanation:
Two unbiased dice are thrown Dice has 1 to 6 numbers
Total Possible out comes = 6 * 6 = 36
(i) the first die shows 6.
(6 , 1) (6 , 2) (6 , 3)( 6. 4)(6 , 5) (6 , 6)
Probability = 6/36 = 1/6
(ii) the total of the numbers on the dice is 8
2 + 6 = 8 , 3 + 5 = 8 , 4 + 4 = 8 , 5 + 3 = 8 6 + 2 = 8
Probability = 5/36
(iii) both the dice show the same number
11 , 22 , 33 , 44 , 55 , 66
Probability = 6/36 = 1/6
(iv) the total of the numbers on the dice is greater than 8.
Total 9 , 10 , 11 , 12
36 , 45 , 46 , 54 , 55 , 56 , 63 , 64 , 65 , 66
Probability = 10/36 = 5/18
(v) the total of the numbers on the dice is 13
maximum sum possible = 12 (6 + 6)
Probability = 0
(vi) the total of the numbers on the dice is any number from 2 and 12, both inclusive
Sum will lies between 2 To 12 only
Hence probability = 1
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