Math, asked by gk0316638, 6 months ago

a box contains 90dics which are numbered from 1to 90.if one disc is drawn at random from the box ,find the probability that it bears a number divisibleby 5

Answers

Answered by saxenalavi422
0

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

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