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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


If 3 baskets are made, what was the probability that Laura made the shots?

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

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

1 Answer 1

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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.
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  • $\begingroup$ Thank you! For Bayes rule, this is correct setup? P(make 3 of 5 shots | 1,2, or 3 dice roll) $\endgroup$
    – user493886
    Oct 22, 2017 at 21:45
  • $\begingroup$ you want to compute $P(1,2, \text{ or } 3 \text{ dice roll}| \text{make } 3 \text{ of } 5 \text{ shots })$ $\endgroup$
    – Siong Thye Goh
    Oct 22, 2017 at 21:48

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