Tom is throwing darts at a target. in his last 30 throws, tom has hit the target 20 times. you want to know the estimated probability that exactly one out of the next three throws does not hit the target.

Step-by-step explanation:

In his last 30 throws, Tom hit the target 20 times and missed 10 times.

p = probability he misses the target = ⅓

q = probability he hits the target = ⅔

Using binomial probability:

P = nCr (p)^r (q)^(n-r)

Given n = 3, r = 1, p = ⅓, and q = ⅔:

P = ₃C₁ (⅓)¹ (⅔)³⁻¹

P = (3) (⅓) (⅔)²

P = 4/9

There is a 4/9 probability that he misses exactly 1 of his next 3 throws.

if he has no friends he cant give them out to his friends

step-by-step explanation:

no friends = cant give anything out

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9