A die is thrown once. Find the probability of getting
(i)    an even number.
(ii)    a number greater than 3.
(iii)    a number between 3 and 6.

Solution

If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒ n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3
Therefore, P(E) = $fraction numerator straight n left parenthesis straight F right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 3 over 6 equals 1 half$

(ii) Let F be the favourable outcomes of getting a number greater than 3, then
F = { 4, 5, 6 }
⇒ n( F) = 3
Therefore, P(E) = $fraction numerator straight n left parenthesis straight F right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 3 over 6 equals 1 half$

(iii) Let G be the favourable outcomes of getting a number between 3 and 6.
G = { 4, 5 }
⇒ n(G) = 2
Therefore, P(G) = $fraction numerator straight n left parenthesis straight G right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 2 over 6 equals 1 third.$

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Real Numbers

A die is thrown once. Find the probability of getting
(i)    an even number.
(ii)    a number greater than 3.
(iii)    a number between 3 and 6.

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