Question

A box contains discs which are numbered between 0 to 100.If one disc is taken at
random from the box, find the probability that it bears (i) a perfect square, (ii) a number
divisible by 5.
plzz answer fast

Answer 1

Answer:

Total outcomes = 98(Between 0 to 100 i.e. 1-99

Step-by-step explanation:

(A) --> A perfect square  =  1, 4, 9, 16, 25, 36, 49, 64, 81

P(A) = 9/98

(B) --> No. divisible by 5  =  5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95

P(B) = 19/98

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Answer 2

▼△▼△▼△GIVEN INFORMATION▼△▼△▼

Box have discs numbered between 0 and 100

means numbers are marked as

1, 2 ,3 ,4 ............................. 99

Let us consider it an A.P. where

first term, a = 1; common difference, d = 1 ;

last term, l = 99 ; no. of terms, n = ?

l = a+(n-1) d

99 = 1+(n-1) (1)

99 = 1+ n - 1

n = 99

It means there are 99 discs in the box

hence, total no. of observations = 99

(You could easily find no. of terms when common difference is 1 in the AP of natural numbers by subtracting first term by last term and then adding 1 ...,I. e. here , n=(99-1 ) +1= 99)

◆◇◆◇SOLUTION OF PART (I) ◆◇◆◇◆◇◆

Perfect square numbers from the discs numbered between 0 & 100 are : 1, 4, 9, 16, 25, 36, 49, 64, 81

No. of perfect square numbers = 9

probability(no.being perfect square) = 9/99 = 1/11

◆◇◆◇◆◇SOLUTION OF PART(II) ◆◇◆◇

numbers between 0 and 100 divisible by 5 are:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95

No. of observations divisible by 5 = 19

probability(no. divisible by 5) = 19/99

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