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`        Numbers are selected at random one at a time from the two digit numbers 00,01,02,03,....99 with replacement .an event  E occurs if and only if the product of the two digits of a selected number is 18. If four numbers are selected  find probability that the event E occurs at least 3 times`
11 months ago

## Answers : (1)

Arun
22540 Points
```							Please find the answer to your questionThe 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 = 24/25From Binomial distributionP (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
```
11 months ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions