A baseball player's batting average is 0.310. If a given game he bats four times, what is the probability that he will get:
a) 4 hits?
b) at most one hit?
c) at least one?
My thoughts:
a) 4P1 x (.310)^1 x (1-0.310)^3 + ... + 4P4 x (0.310)^4 x (1-0310)^0
b) 4P1 x (0,310)^1 x (1-0.310)^3
c) is c) the same as a)
$(a)\quad$ The probability of obtaining a hit when batting once is $0.31$. So the probability of obtaining $4$ hits when batting $4$ times is
$$(a)\quad \,(0.31)\times (0.31)\times (0.31) \times (0.31) = (0.31)^4$$
$(b)\quad$ The probability of making at most one hit is the probability of obtaining (exactly one hit) OR (no hits). Let's see what this means, in terms of outcomes:
hit not not not Probability: $\quad (0.31)(1 - 0.31)(1 - 0.31)(1 - 0.31) = (0.31)(1 - 0.31)^3$
not hit not not Probability: $\quad (0.31)(1 - 0.31)^3$
not not hit not Probability: $\quad (0.31)(1 - 0.31)^3$
not not not hit Probability: $\quad (0.31)(1 - 0.31)^3$
not not not not Probability: $\quad (0.31)^4$
Probability of obtaining at most one hit: we add each of the above
$$(b)\quad 4\cdot (0.31)(1-0.31)^3 + (0.31)^4 = 4\cdot(0.31)(0.69)^3 + (0.31)^4$$
$(c)$ This is not the same as $(a)$. We compute the probability of getting no hits, and subtract from $1$ this probability of getting no hits. That difference will give us the probability of getting at least one hit.
Probability of getting no hits $= (1 - 0.31)^4 = (0.69)^4$.
(c) So the probability of getting at least one hit is equal to $\;1 - \left((0.69)^4\right)$
