9. The probability that a two digit number selected at random will be a multiple of 3 and not a multiple of 5 is​

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​”

1. 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

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2. 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

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