# Binomial Probability with Condition

Do not have answers to this question
Please let me know if there are mistakes

Three basketball players are practicing shooting free throws.

• Laura typically makes 80% of her shots
• Alice typically makes 50% of her shots
• Shelly typically makes 70% of her shots

a) In the first drill, each girl takes 5 free throws. You may assume that the probabilities of making each throw are independent. What is the probability that Laura will make exactly 3 baskets?

Attempt:

${5 \choose 3}{0.8}^3{0.2}^2 = 0.2048$

b) What is the probability that Alice will make an odd # of baskets?

Attempt:

\begin{align} P(odd) &= 1 - P(even) \\ \ &= 1 - P(2 or 4) \\ \ &= 1 - P(2) + P(4) - P(2 and 4) \\ \ &= 1 - {5 \choose 2}{0.5}^2{0.5}^3 + {5 \choose 4}{0.5}^4{0.5}^1 - P(2 and 4) \\ \ &= 1 - 0.3125 + 0.15625 - (0.3125 * 0.15625) \\ \ &= 1 - 0.41992 \\ \ &= 0.5800 \end{align}

c) In the next drill, the coach rolls a 6-sided die.

• If 1,2 or 3 comes up, Laura takes 5 free throws

• If 4 or 5 comes up, Alice takes 5 free throws

• If 6 comes up Shelly takes 5 free throws

Attempt:

\begin{align} P(Laura AND3 Basket) &= {5 \choose 3}{0.8}^3{0.2}^2 * {3\over 6} \\ \ &= 0.1024 \end{align}

Thank you

• It is not possible for Alice to make both $2$ and $4$ baskets. She either makes $2$ or she makes $4$. Oct 22, 2017 at 21:32
• ... or she could make $0$. An intuitively simpler answer to (b) could be $P(A=1)+P(A=3)+P(A=5)$ which is equal to $1-(P(A=0)+P(A=2)+P(A=4))$ Oct 22, 2017 at 21:34
• Thank you Henry and N. F. Taussig. I did P(1) + P(3) + P(5) and get 0.5
– user493886
Oct 22, 2017 at 21:38

Part b:

• There is another even number, $0$.
• As mentioned in the comment, we can't make $2$ and $4$ basket simultaneously
• Be careful when you perform expansion $-(A+B)=-A\color{red}{-}B$

Part c:

• We are interested in conditional probability, use Bayes rule.
• Thank you! For Bayes rule, this is correct setup? P(make 3 of 5 shots | 1,2, or 3 dice roll)
– user493886
Oct 22, 2017 at 21:45
• you want to compute $P(1,2, \text{ or } 3 \text{ dice roll}| \text{make } 3 \text{ of } 5 \text{ shots })$ Oct 22, 2017 at 21:48