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

0.2946

Step-by-step explanation:

Number of tosses, n = 200

P(obtaining a 5), p = 1/6

q = 1 - p = 5/6

Normal approximation for binomial distribution

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 200 x 1/6

= 33.33

Standard deviation = √npq

= √(200(1/6)(5/6) )

= 5.27

P(at most 30 fives) = P(X ≤ 30)

= P(Z < (30.5 - 33.33)/5.27) (continuity correction of 0.5 is added to 30)

= P(Z < -0.54)

= 0.2946