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Q.

Box A contains 25 slips of which 19 are marked ₹1 and other are marked ₹5 each. Box B contains 50 slips of which 45 are marked ₹1 each and other are marked ₹13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than ₹1?


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a

11 75

b

12 75

c

13 75

d

14 75  

answer is A.

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Detailed Solution

Given that,
Total number of slips in box A = 25.
Total number of slips in box B = 50.
No. of slips marked Rs.1,
In the box A = 19,
In the box B = 45.
We know,
P(E)= n(E) n(S)  and P( E ¯ )=1−P(E) ,
where P(E) and P( E ¯ ) are the probability of occurrence and non-occurrence of event E.
After pouring slips of both boxes in a third box,
Total number of slips in the third box =n S = 25+50 =75
Total number of slips marked Rs.1 in the third box =n E
=19+45=64 Probability of getting Rs.1 marked slip =P(E) .
P(E)= n(E) n(S) = 64 75
Now, the probability of getting slips other than Rs.1 = 1−P(E) .
P( E ¯ )=1−P(E) =1− 64 75 = 75−64 75 = 11 75 Hence, the probability that the slip is marked other than Rs. 1 is 11 75 .
Therefore, option 1 is correct.
 
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