2 thoughts on “9. The probability that a two digit number selected at random will be a multiple of 3 and not a multiple of 5 is”

Step-by-step explanation:

There are 90 two digit numbers(99–9). Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5. Therefore, the favorable cases are 30–6=24. Hence, the required probability is 24/90 = 4/15

Step-by-step explanation:There are 90 two digit numbers(99–9). Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5. Therefore, the favorable cases are 30–6=24. Hence, the required probability is 24/90 = 4/15

Answer:24/90 = 4/15 is the answer

Step-by-step explanation:All two digit numbers are:

{10,11, 12, 13, 14, 15, …. 99}

= (99-10) + 1 = 90

Digits divisible by 3 are: {12, 15, 18, … 99}

As they form an AP,

12 + (n-1)×3 = 99

=> 3(n – 1) = 87

=> n – 1 = 29

=> n = 30

Digits divisible by both 3 and 5(i.e. divisible by 15) {15, 30, 45, 60, 75, 90} = 6

Therefore, Numbers divisible by 3 but not 5 = 30 – 6

= 24

Probability = 24/90