Q. 12 A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at
random from the box. Find the probability that it bears.
(IH) A two digit number (II) A perfect square number (iii) A number divisible by 5.
Answers
Answer:
Let E be the event of drawing a disc from the box of discs numbered from 1 to 90
Two digit numbers from 1 to 90=10,11,12,.....,90
No. of favorable outcomes=81
Total no. of possible outcomes =90
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
90
81
=
10
9
Therefore, the probability of a two digit numbered disc from the box of discs numbered from 1 to 90=
10
9
Solution(ii):
Let F be the event of drawing a perfect square numbered disc from the box of discs numbered from 1 to 90
Perfect square numbers from 1 to 90=1,4,9,16,25,36,49,64,81
No. of favorable outcomes=9
Total no. of possible outcomes =90
We know that, Probability P(F) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
90
9
=
10
1
Therefore, the probability of drawing a perfect square numbered disc from the box of discs numbered from 1 to 90=
10
1
Solution(iii):
Let G be the event of drawing a disc with a number divisible by 5 from the box of discs numbered from 1 to 90
Numbers divisible by 5 from 1 to 90=5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90
No. of favorable outcomes=18
Total no. of possible outcomes =90
We know that, Probability P(G) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
90
18
=
5
1
Therefore, the probability of drawing a disc with a number divisible by 5 from the box of discs numbered from 1 to 90=
5
1