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Q.

Cards with numbers 2 to 101 are placed in a box. A card is selected at random. What is the probability that the card has an even number?

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a

\frac{1}{2}

b

\frac{1}{9}

c

0

d

1

answer is A.

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Detailed Solution

Given that cards with numbers 2 to 101 are placed in a box.
A card is selected at random.
Let us assume that P(E) is the probability of an event “E”, n(S) is the total number of events in the sample space and n(E) is the number of favorable outcomes.
The total number of outcomes with numbers 2 to 101 is,

n S = 100

.
The card which has an even number from 2 to 101 would be {2,4,6, ........100}.
The series {2,4,6, ........100} is an A.P series which have the value,

a = 2

,

l = 100

and the common difference is,

d = 2

.
Substitute 2 for a and 2 for d and 100 for l in

l = a + n − 1 d

to obtain the number of terms. So,

l = a + n − 1 d ⇒ 100 = 2 + n − 1 2 ⇒ 100 − 2 = 2 n − 2 ⇒ 2 n = 100 ⇒ n = 50

So, the number of cards which have even numbers of cards are,

n E = 50

.
The probability that the card has an even number,

P E = n E n S P E = 50100 P E = 12

Therefore, 1 is the correct option.

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