what is this question to do with E
Numbers are selected at random one at a time,from the numbers 00,01,02,...99 with replacement.An event E occurs if and only if the product of the two digits of a selected number is 18.If 4 numbers are selected ,then the probability that E occurs at least 3 times,is
a)97/390625
b)68/390625
c)72/390625
d)79/390625
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4 Answers
this question a confusing problem
hint : probability of picking a number whose product of 2 digits is 18=4/100=1/25
btw the answer is A!
just give it a little attempt.. (it is not too difficult either!)
Answer = probabl of getting exactly 3 + probab of getting exactly 4
probability of picking a number whose product of 2 digits is 18=4/100=1/25
Probability of exactly 3 = 4C3 (1/25)3 (24/25)
Probability of exactly 4 = 4C4 (1/25)4
Take the sum.. u get the answer :)