Real Numbers

Question

A die is thrown once, find the probability of getting a prime number.

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 outcome of getting a prime number then
E = 12, 3, 5}
⇒ irt(E) = 3
Therefore, P(E) = $fraction numerator straight n left parenthesis straight E right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 3 over 6 equals 1 half$

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Real Numbers

A die is thrown once, find the probability of getting a prime number.

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