A box containing cards numbered between 0 and 100 are shuffled and a card is picked at random. Find the probability of getting a card which is:
(a) divisible by 6.
(b) not divisible by 6.
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The cards are numbered from 0 to 100.
Total number of cards = 101 (as there is no. zero card also)
(a) Favorable outcome = Getting a card having a number which is divisible by 6.
{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}
Probability $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{101}$.
Hence, probability of getting a card which is divisible by 6 = $\frac{17}{101}$.
(b) Favorable outcome = Getting a card having a number which is not divisible by 6.
P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)
= $1 - \frac{17}{101}$
= $\frac{101 - 17}{101}$
= $\frac{84}{101}$.
Hence, probability of getting a card which is not divisible by 6 = $\frac{84}{101}$.