Question

2. A box contains discs which are numbered from 2 to 101. If one disc is drawn at random from
the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
(iv) a prime number less than 100

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Answer 1

Answer:

Step-by-step explanation:

The total number of outcomes = 101 - 2 + 1 = 100

a) Two digit numbers are from 10 to 99 which are 90 in number.

=> P(getting a two digit number) = 90/100 = 9/10.

b) Perfect squares are  4, 9, 16, 25, 36, 49, 64, 81, 100

=> P(getting a perfect square) = 9/100

c) Numbers divisible by 5 are 5, 10, 15,20,25,30,35,40,45,50,55, 60, 65, 70, 75,80,85,90,95, 100, which are 20 in number.

=> P(getting a number divisible by 5) = 20/100 = 1/5.

d) Prime number less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97  which are 25 in number.

=> P(Prime number less than 100) = 25/100 = 1/4.

Answer 2

a+(n-1)*d =an

2+(n-1)*1 =101

n-1 = 99

n = 100

for two digit number

100-9

=91

probability= 91/