Math, asked by jain10harsh29, 1 year ago

. If the probability of hitting a target by a shooter in any
shot is 1/3 , then the minimum no. of
independent shots at the target required by him so that
the probability of hitting the target atleast once is
greater than 5/6, is
(a) 6 (b) 3 ( c) 5(d) 4

Answers

Answered by Fatimakincsem
1

The probability of hitting the target at least once is  greater than 5/6 is 5.

Step-by-step explanation:

Let the no. of attempts be = n

Probability of success = 1/3

Probability of failure = 2/3

Probability of success at least once =  1 - probability of no success

                                         = 1 - nCn (2/3)^n

                                         = 1 - ( 2/3)^n

Now

1 - ( 2/3) > 5/6

So n = 5

Thus the probability of hitting the target at least once is  greater than 5/6 is 5.

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What is the probability of getting a pair of numbers whose ratio is in simplest form in 2 dice thrown?

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