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Numbers are selected at random, one at a time, from the two digit numbers 00, 01, 02, . . . . . . . . . . . . . . ., 99 with replacement. An even E occurs if only if the product of the two digits of a selected numbers is 18. If four numbers are selected, find probability that the event E occurs at least 3 times.Numbers are selected at random, one at a time, from the two digit numbers 00, 01, 02, . . . . . . . . . . . . . . ., 99 with replacement. An even E occurs if only if the product of the two digits of a selected numbers is 18. If four numbers are selected, find probability that the event E occurs at least 3 times.

Numbers are selected at random, one at a time, from the two digit numbers 00, 01, 02, . . . . . . . . . . . . . . ., 99 with replacement. An even E occurs if only if the product of the two digits of a selected numbers is 18. If four numbers are selected, find probability that the event E occurs at least 3 times.Numbers are selected at random, one at a time, from the two digit numbers 00, 01, 02, . . . . . . . . . . . . . . ., 99 with replacement. An even E occurs if only if the product of the two digits of a selected numbers is 18. If four numbers are selected, find probability that the event E occurs at least 3 times.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
The given numbers are 00, 01, 02, . . . . . . . . . . . . . . . . 99. These are total 100 numbers, out of which the numbers, the product of whose digits is 18, are 29, 36, 63 and 92.
∴ p = P (E) = 4/100 = 1/25 ⇒ q = 1 – p = 24/25
From Binomial distribution
P (E occurring at least 3 times) = P (E occurring 3 times) + P (E occurring 4 times)
4C3 p3 q + 4C4 p4 = 4 x (1/25)3 (24/25) + (1/25)4 = 97/(25)4

Thanks
Aditi Chauhan
askIITians Faculty

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