Question

There are 30 cards, of 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

Answer 1

SOLUTION :  

Given : Number of cards in a bag = 30

Total number of outcomes = 30

Let E = Event of getting a number not divisible by 3

Numbers which are not divisible by 3 are = 1, 2, 4, 5, 10, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29

Number of outcome favourable to E = 10

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

P(E) = 20/30 = 2/3

Hence, the required probability of getting a number not divisible by 3, P(E) = ⅔  

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Answer 2

Hi there !
Here's the answer:

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Given,
Total No. of Cards = 30
These 30 cards are numbered from 1 - 30
A card is drawn at random

Let S be the Sample Space
n(S) - No. of ways of drawing a card from 30 cards
n(S) = 30C1 = 30

Let E be the Event that the drawn card contains the number which is not a multiple of 3

E' be Event that drawn card contains number which is a multiple of 3

From 1- 30,
No. of Multiplies of 3 = 30/3 = 10
(which are 3, 6, 9,......., 30)

No. of favourable outcomes satisfying occurrences of Event E', n(E') = 10

n(E) + n(E') = n(S)
=> n(E) = 30 - 10 = 20

No. of favourable outcomes satisfying occurrences of Event E, n(E) = 20
[E = { 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29} ]

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

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

P(E) = $\frac{20}{30}$

P(E) = $\frac{2}{3}$

•°• Required Probability = $\frac{2}{3}$

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