A die is rolled, then the probability that a number 1 or 6 may appear is
(a) 2/3
(b) 5/6
(c) 1/3
(d) 1/2

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Solution

(c) 1/3

Total number of sample space, S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let A be the event of getting the number 1 or 6.
A = {1, 6}
i.e. n(A) = 2
Hence, required probability = $\frac{n (A)}{n (S)} = \frac{2}{6} = \frac{1}{3}$

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A die is rolled, then the probability that a number 1 or 6 may appear is (a) 2/3 (b) 5/6 (c) 1/3 (d) 1/2

(c) 1/3 Total number of sample space, S = {1, 2, 3, 4, 5, 6} ∴ n(S) = 6 Let A be the event of getting the number 1 or 6. A = {1, 6} i.e. n(A) = 2 Hence, required probability = nAnS=26=13

A die is rolled, then the probability that a number 1 or 6 may appear is
(a) 2/3
(b) 5/6
(c) 1/3
(d) 1/2

(c) 1/3

Total number of sample space, S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let A be the event of getting the number 1 or 6.
A = {1, 6}
i.e. n(A) = 2
Hence, required probability = $\frac{n (A)}{n (S)} = \frac{2}{6} = \frac{1}{3}$