A hat contains 80 slips of paper. Of these, 20 are green and labeled from 1 to 2

Question

A hat contains 80 slips of paper. Of these, 20 are green and labeled from 1 to 20, 10 are red and labeled 1 to 10, 40 are yellow are labeled 1 to 40, 10 are blue and labeled 1 to 10. These are well mixed so that each paper is equaly likely to be chosen in one draw from the hat. Find the probability that the paper is

a) green or red
b) numbered 1,2,3,4, or 5
c) red or yellow, and also numbered 1,2,3, or 4 d) numbered 5, 15, 25, 35
e) yellow if it is labeled either 19 or 20
f) yellow if it is labeled either 3 or 4?

Explanation / Answer

a) P(Green or red) = P(Green) + P(Red) - P(Green and Red)

Since

There is no paper with both green and red color.

P(Green and red) = 0

Hence,

P(Green or Red)

= P(Green) + P(Red)

= 20/80 + 10/80

= 3/8

b) P(Paper numbered 1,2,3,4 or 5)

= P(1) + P(2) + P(3) + P(4) + P(5)

There are 1 green, 1 red, 1 yellow and 1 blue paper marked 1. So 1 can be drawn in 4 ways. Similarly, 2, 3, 4 and 5 also can be drawn in 4 ways. Hence,

P(Paper numbered 1, 2, 3, 4 or 5)

= 4/80 + 4/80 + 4/80 + 4/80 + 4/80

= 20/80

= 1/4

c) P(Red numbered 1, 2, 3 or 4 or Yellow numbered 1, 2, 3 or 4)

= P(Red numbered 1,2,3 or 4) + P(Yellow numbered 1,2,3 or 4)

= 4/80 + 4/80

= 1/10

d) P(Numbered 5, 15, 25, 35)

= P(Numbered 5) + P(Numbered 15) + P(Numbered 25) + P(Numbered 35)

= 4/80 + 2/80 + 1/80 + 1/80

= 8/80

= 1/10

e) P(Yellow | Lebeled 19 or 20)

= P(Yellow and Labeled 19 or 20) / P(Labeled 19 or 20)

= (2/80) / (4/80)

= 2/4

= 1/2

f)

P(Yellow | Lebeled 3 or 4)

= P(Yellow and Labeled 3 or 4) / P(Labeled 3 or 4)

= (2/80) / (8/80)

= 2/8

= 1/4

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