Question

A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box. Find the probability that the number on the drawn card is a prime number.

Answer 1

SOLUTION :  

Given : Cards marked with numbers 3, 5, 7, 9 …...35, 37

These Numbers are in A.P  

Here, a = 3 , d = 5 - 3 = 2 , an = 37

Let n = number of terms  

an = a + (n - 1) d

37 = 3 + (n - 1) × 2

37 = 3 + (2n - 2)

37 = 3 - 2 + 2n  

37 = 1 + 2n  

2n = 37 - 1

2n = 36

n = 36/2  

n = 18

Total number of outcomes = 18

Let E = Event of getting a prime number  

Numbers which are prime  = 3, 5, 7,11, 13, 17, 19, 23, 29, 31 & 37

Number of outcome favourable to E = 11

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 11/18  

Hence, the required probability of getting a prime number , P(E) = 11/18  .

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Answer 2

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER:

¶ In first 'n' Natural Numbers,

- there are $\frac{n}{2}$ odd Numbers ; When n-odd

- there are $\frac{n+1}{2}$ odd numbers ; when n- even.

¶ Probability of Occurrence of an Event = $\frac{No.\: of\: Favourable\: Outcomes}{Total\: No.\: of\: Outcomes}$

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•

¶ SOLUTION:

Given,
A box contains cards numbered 3, 5, 7, 9, ………, 35, 37

All are odd No.s from 3 to 37

In 1-37,
there are $\frac{37+1}{2}=19$ odd Numbers

But 1 is excluded in the given cards

•°•
Total No. of cards in the box = 19 - 1 = 18

A card is drawn at random

Let S be the Sample Space
n(S) - No. of ways of drawing a card from 18 cards

n(S) = 18C1 = 18

Let E be the event that the drawn card contains a prime number

E = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}

n(E) - No. of favourable outcomes for occurrence of Event E
= No. of elements in set E

n(E) = 11

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

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

$P(E) = \frac{11}{18}$

•°• Required Probability = $\frac{11}{18}$

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