Question

• Finding probability for various situations like when a dice is rolled 100 times the probability
of getting (i) even, (ii) odd, (ii) prime etc.

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Answer 1

(i) P(getting an even number)$=(\dfrac{1}{2})^{100}$

(ii) P(getting an odd number)$=(\dfrac{1}{2})^{100}$

(ii) P(getting a prime number)$=(\dfrac{1}{2})^{100}$

Step-by-step explanation:

When a dice is rolled one times.

All possible outcomes are:

1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

(i) even

The even numbers are 2, 4 and 6.

∴ Favourable outcomes = 3

∴P(getting an even number) = $\dfrac{3}{6} =\dfrac{1}{2}$

(ii) odd

The odd numbers are 1, 3 and 5.

∴ Favourable outcomes = 3

∴P(getting an odd number) $= \dfrac{3}{6} =\dfrac{1}{2}$

(iii) prime

The prime numbers are 2, 3 and 5.

∴ Favourable outcomes = 3

∴P(getting a prime number) $= \dfrac{3}{6} =\dfrac{1}{2}$

Therefore, when a dice is rolled 100 times.

(i) P(getting an even number) $=(\dfrac{1}{2})^{100}$

(ii) P(getting an odd number)$=(\dfrac{1}{2})^{100}$

(iii) P(getting a prime number)$=(\dfrac{1}{2})^{100}$