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There are 30 cards, of the same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.

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23

13

34

16

answer is D.

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Detailed Solution

Given that cards bearing numbers 1 to 30 are kept in a bag.
We have to find the probability that when a card is drawn, it is bearing a number not divisible by 3.
Probability of an event E is defined as:

P (E) = \frac{Number of favorable outcomes}{Total number of possible outcomes}

Since the cards are bearing numbers 1 to 30 therefore, the total number of possible outcomes is 30.
Cards bearing numbers not divisible by 3 are:

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29

Hence, the number of favorable outcomes is 20.
Let

E

be the event of getting a card bearing a number not divisible by 3
Now, the probability of getting a card bearing a number not divisible by 3 is:

P (E) = 2030 ⇒ P (E) = 23

The probability of getting a card bearing a number not divisible by 3 is

23

.
Therefore, option 4 is correct.

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