Question

An integer is chosen at random from the first 100 positive integers. Then the probability that the number chosen is divisible by 3 and 4 equals

Answer 1

Answer:

We have,

x+

x

100

>50

⟹x

2

+100>50x

⟹x

2

−50x+100>0

⟹(x−25)

2

−525>0

⟹x−25>

525

or x−25<

525

⟹x<25−

525

or 25+

525

As x is positive and

525

=22.9

We must have x≤2 or x≥48

Thus 2 + 53 = 55 favorable cases.

Probability =

100

55

=

20

11

Answer 2

The first 100 positive integers are numbers from 0 to 99 both inclusive.
A number divisible by 3 and 4, should be divisible by 12
No. Of favorable outcomes
= no. Of Multiples of 12 from 0 to 9
= 8 (since 12 x 8 = 96 and all other numbers are greater than 99)
Probability
= no. Of favorable outcomes / total no. Of outcomes
= 8/100 = 0.08

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