Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives

Solution:

Given, a fair die is tossed 200 times.

We have to find the probability to obtain at most 30 fives. I.e.x <= 30

P(x <= 30)

We can find the probability by using binomial distribution,

Here, n = 200

p = probability of success

When a die is tossed, the probability of getting a five is 1/6

p = 1/6

q = 1 - p

q = 1 - (1/6)

q = 5/6

By formula,

Mean = np

np = 200(1/6)

Mean = 33.33

Standard deviation = √(npq)

Standard deviation = √(200)(1/6)(5/6)

= √(1000/36)

On simplification,

Standard deviation = √[(100)(10)/36]

= √27.778

= 5.270

By using Z formula,

Z = (x - mean)/standard deviation

Z = (30 - 33.33)/5.270

Z = -3.33 / 5.270

Z = -0.632

P(Z < -0.632) = 0.2643

P(x<=30) = 0.2643

Therefore, the probability of obtaining at most 30 fives is 0.2643.

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Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives

Summary:

When a fair die is tossed 200 times, the probability of obtaining at most 30 fives is 0.2643.