Question

Mark the correct alternative in each of the following: A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is
(a)$\frac{2}{3}$
(b)$\frac{1}{6}$
(c)$\frac{1}{3}$
(d)$\frac{11}{30}$

Answer 1

SOLUTION :

The correct option is (C)   : ⅓

Given : A number is selected from 1 to 30

Total number of possible outcome = 30

Let E = Event of getting a prime number

Prime number from 1 to 30  are = 2, 3,5,7, 11,13,17, 19, 23, 29

Number of outcomes  favorable to E = 10

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

P(E) = 10/30 = ⅓  

Hence, the Probability of getting a  prime number from 1 to  30 ,P(E)  is 1/3.

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

Hi there !
Here's the answer:

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Given,
A Number is selected at Random from 1 to 30

Let S be Sample Space
n(S) - No. of ways of selecting a card from 30 cards

n(S) = 30C1 = 30

Let E be the Event that the Number selected is a Prime Number

E = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

n(E)- No. of favorable outcomes for Occurrence of Event E.

n(E) = 10

Probability = $\dfrac{No.\: of\: Favourable\: Outcomes}{Total\: No.\: of\: Outcomes}$

$P(E) = \frac{n(E)}{n(S)}$

$P(E) = \frac{10}{30}= \frac{1}{3}$

•°• Required Probability = $\frac{1}{3}$



This answer exists in option (c)

•°• Option (c) is correct

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