A packet contains some blue, red and green pens such that the probability of picking a blue pen is 2

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13 Apr 2022 Probability

A packet contains some blue, red and green pens such that the probability of picking a blue pen is 2/9 and probability of picking a green pen is 4/9. If number of red pens is 15 and if all the pens are numbered starting from 1, 2, 3, ... and so on, then what is the probability of getting one pen numbered as multiple of 7 or 11 ?

OPtion

1) 8/15
2) 4/15
3) 4/9
4) 1/5
5) 2/9
6) 7/20
7) 5/9
8) 1/3
9) 2/5
10) None of these

Solution

Given, P(Blue)=2/9, P(Green)=4/9 and number of red pens=15
As we know, P(Blue) + P(Red) + P(Green) = 1
P(Red) = 1 - P(Blue) - P(Green) = 1 - (2/9) - 4/9 = 3/9 = 1/3
As number of red pens = 15 = 1/3 of total pens
.'. Total number of pens=15*3=45 ⇒ Pens are numbered from 1 to 45
Numbers that are multiple of 7 = 7, 14, 21, 28, 35, 42
And numbers that are multiple of 11 = 11, 22, 33, 44
So, favourable outcomes = 6+4 = 10
So, probability of getting pen of multiple 7 or 11 = 10/45 = 2/9

Correct Option: 5 Best Solution (2)

A.P.Singh   2 years AGO

Ratios of blue, red, green pens are 2:3:4.
Number of red pens=15
Hence total number of pens = 15*9/3=45
Multiples of 7 or 11 till 45 are (7,14,21,28,35,42) & ( 11,22,33,44) i.e. total 10 numbers.
so required probability =10/45=2/9

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Mayank Kandoi   2 years AGO

Probability of getting red pen = 1 - 2/9 - 4/9 = 3/9 = 1/3
Total red pens = 15
Let total pens = x
Then, x/3 = 15 which means x = 45

Favorable outcomes = 7,11,14,21,22,28,33,35,42,44
Total outcomes = 45
Probability = 10/45 = 2/9

Option 5

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