Two numbers are selected at random from 123.....100 and are multiplied then the probability correct to two places of decimals that the product thus obtained is divisible by 3 is ????
Answer (1)
5th Jul, 2021
Greetings of the day dear aspirant
Total cases = 100 C 2
Favorable cases = 33 C 1 . 67 C 1 + 33 C 2
(We know that from 1 to 100, 33 numbers are exactly divisible by 3, such that if any of these numbers are multiplied by remaining 67 numbers, their product should be divisible by 3, or any two numbers from 33 numbers are multiplied, their product should be divisible by 3)
Hence, the required probability = ( 33 C 1 . 67 C 1 + 33 C 2 )/ 100 C 2
= [(33. 67) +528]/ 4950
=(2211+528)/4950
= 2739/4950
= 0.55.
Hope this helps!!
All the best for your future
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