Real Numbers

Question

A die is thrown once. What is the probability of getting a number 5 or 6?

Solution

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

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Real Numbers

A die is thrown once. What is the probability of getting a number 5 or 6?

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