Question

Mark the correct alternative in each of the following: A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23, is
(a)$\frac{7}{90}$
(b)$\frac{10}{90}$
(c)$\frac{4}{45}$
(d)$\frac{9}{89}$

Answer 1

SOLUTION :  

The correct option is (c) : 4/45

Given : There are 90 discs numbered from 1 to 90  are placed in a box.

Total number of outcomes = 90

Let E = Event of getting a prime number less than 23  

Prime number less than 23 are : 2, 3,5,7, 11,13,17, 19

Number of outcomes  favorable to E = 8

Probability ,P(E) = Number of favourable outcomes / total number of outcomes

P(E) = 8/90 = 4/45

Hence, the Probability of getting a prime number less than 23 ,P(E)  is 4/45.

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Answer 2

correct answer is) 4/45

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