Question 1186983: 5) Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer in a particular store. The six possible outcomes and their probabilities are given in the table.
a) Let event A = grip size of in.
Find P(A).
Interpret this probability.
b) Find P(AC),
c) What is the probability that the racket purchased has an oversize head (event B)?
d) What is the probability that grip size is at least in.?
e) What is the probability that the grip size is inch and the head is oversized?
f) What is the probability that the grip size is inch or the head is oversized?
g) If the racket has a midsized head, what is the probability that the grip size is inch?
h) Are the events the grip size of inch and oversized head independent?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this probability problem. I'll need the table of outcomes and probabilities to give you specific numerical answers. However, I will show you the general method. *Please provide the table so I can complete the calculations.*
**General Method and Explanation:**
Let's represent the different head sizes as M (midsized), O (oversized), and E (extra-oversized), and the grip sizes as S (small), M (medium), and L (large). The table would look something like this (but with actual probabilities):
| Outcome (Head, Grip) | Probability |
|---|---|
| (M, S) | P(M, S) |
| (M, M) | P(M, M) |
| (M, L) | P(M, L) |
| (O, S) | P(O, S) |
| (O, M) | P(O, M) |
| (O, L) | P(O, L) |
| (E, S) | P(E,S) |
| (E,M) | P(E,M) |
| (E,L) | P(E,L) |
*Remember: The sum of all probabilities must equal 1.*
**a) P(A): Grip size of 1/4 inch**
Event A corresponds to all outcomes where the grip size is 1/4 inch (which I'm assuming is represented by "S" for small in my example).
P(A) = P(M, S) + P(O, S) + P(E,S)
*Interpretation:* P(A) is the probability that a randomly selected customer purchased a racket with a small (1/4 inch) grip.
**b) P(Aᶜ):**
P(Aᶜ) = 1 - P(A) (This is the complement of event A; the probability that the grip size is *not* 1/4 inch).
**c) P(B): Oversized head**
Event B corresponds to all outcomes where the head size is oversized (O).
P(B) = P(O, S) + P(O, M) + P(O, L)
**d) Grip size at least 1/2 inch**
"At least 1/2 inch" means medium (M) or large (L) grip sizes.
P(grip ≥ 1/2) = P(M, M) + P(M, L) + P(O, M) + P(O, L) + P(E,M) + P(E,L)
**e) Grip size 1/4 inch and oversized head**
This is the probability of the intersection of events A and B.
P(A ∩ B) = P(O, S)
**f) Grip size 1/4 inch or oversized head**
This is the probability of the union of events A and B.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
**g) Probability of 1/4 inch grip given midsized head**
This is a conditional probability.
P(A | M) = P(A ∩ M) / P(M) = P(M,S) / [P(M,S) + P(M,M) + P(M,L)]
**h) Independence of events**
Events A and B are independent if P(A ∩ B) = P(A) * P(B). Calculate P(A) * P(B) and compare it to the value you got for P(A ∩ B) in part (e). If they are equal, the events are independent. If they are not equal, the events are dependent.
**Provide the table, and I'll calculate the specific probabilities for you!**
Answer by ikleyn(52798)  (Show Source):
|
|
|
