Harry has a batting average of .200 (i. e., a 20 percent chance of a hit each time he's at bat). Scouts for a rival baseball club secretly observe Harry's performance in 12 random times at bat. What is the probability that Harry will not get more than 2 hits

0.9999

Step-by-step explanation:

Probability of a bit whenever he is at bat = 0.20

Probability when not having a hit = 1-0.20

= 0.80

n = 12

P = 0.80

We have a binomial distribution

P(X=x) = nCx*P^x(1-p)^n-x

P(x>2) = 1-p(x<=2)

= 1-[(12C0 * 0.8⁰*(1-0.8)¹²) +(12C1 * 0.8¹ * (1-0.8)¹¹) + (12C2 * 0.8² * (1-0.8)¹⁰)]

= 1-(0.000000004096+0.000000196608+0.000004325376)

= 1-0.00000452608

This is approximately

0.9999

c.)

step-by-step explanation:

y=sec x is the correct answer

~batmans wife dun dun