There are two buckets containing same type of balls of different colours as below:
Bucket A – Red 10%, White 60% and black 30%
Bucket B – Red 10%, White 40% and black 50%
Someone has to select 1 bucket with equal probability and then 1 ball is to be drawn from
the selected bucket. Answer the following questions: (1+1+1+1)
(a) The probability of selecting bucket A is
i) 1/3
ii) 1/4
iii) 1/2
iv) None of the above
(b) The sample spaces for the above problem is
i) {R,B,W}
ii) {(1,R), (1,B), (1,W), (2,R), (2,B), (2,W)}
iii) {(1,R), (2,R)}
iv) {1,2} (c) If E is the event bucket 2 selected and F is the event white ball drawn. Then the value
of P (F/E) is
i) 0.1
ii) 0.05
iii) 0.2
iv) 0.4
(d) If F is the event for white ball drawn then P(F) is
i) 0.6
ii) 0.5
iii) 0.4
iv) None of the above
(e) If white ball is drawn then the probability that it is drawn from 1st bucket is
i) 3/5
ii) 2/5
iii) 4/5
iv) 1/5
Answers
Given : There are two buckets containing same type of balls of different colors as below:
Bucket A – Red 10%, White 60% and black 30%
Bucket B – Red 10%, White 40% and black 50%
To Find : The probability of selecting bucket A
The sample spaces for the above problem is
(c) If E is the event bucket 2 selected and F is the event white ball drawn. Then the value of P (F/E) is
If F is the event for white ball drawn then P(F) is
If white ball is drawn then the probability that it is drawn from 1st bucket is
Solution:
The probability of selecting bucket A = 1/2 as there are 2 buckets
The sample spaces for Selecting bucket is or ( 1 , 2)
The sample spaces for selecting ball
( 1 , R) , (1 , B) , (1 , W) , ( 2, R) , (2 , B) , (2 , W)
) If E is the event bucket 2 selected and F is the event white ball drawn. Then the value of P (F/E) is = 40/100 = 0.4
F is the event for white ball drawn then P(F) = (1/2) (60/100) + (1/2)(40/100)
= 0.5
white ball is drawn then the probability that it is drawn from 1st bucket is
White ball from bucket A = 0.6
White ball from bucket B = 0.4
= (0.6)/(0.6 + 0.4)
= 0.6
= 3/5
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