During basketball practice, Mai attempted 40 free throws and was successful on 25% of them. How many successful free throws did
she make?
2 answers:
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Answer:
a) The probabilities are:
Pr(1) = 1/4
Pr(2) = 1/4
Pr(3) = 1/4
Pr(4) = 1/4
b) Pr(sum of 6 or more on two throw) = 6/16 = 3/8.
c) If the die was fair, it should have six (6) numbers => (1,2,3,4,5,6) and the sample space from two throw will be = 36.
Hence, Pr(sum of 6 or more on two throw) = 26/36 = 13/18.
Step-by-step explanation:
a) Since the probability of a face is proportional to the number of dots (there are 1, 2, 3, and 4 dots on the respective sides), then the occurrence of any outcome in a single throw are equally likely. Hence, there is equal probability of occurrence.
By probability, we say
Pr(x) = number of favorable event/ total number of possible events
Therefore, the total number = 4, and each can occur at equal probability. This is why the answer are:
Pr(1) = 1/4
Pr(2) = 1/4
Pr(3) = 1/4
Pr(4) = 1/4
b) The sample space when the die is thrown twice = = 16 possible outcome.
Among the outcomes, those that will have 6 or more when sum are just six (6). They are:
(2,4), (3,3), (3,4), (4,2), (4,3), and (4,4)
Thus, Pr(sum of 6 or more on two throw) = 6/16 = 3/8.
Just for understanding:
Other outcome from the sample space are:
(1,1), (1,2), (1,3), (1,4)
(2,1), (2,2), (2,3), ___
(3,1), (3,2), ___, ___
(4,1), ___, ___, ___
If you observe, the dash are the ones we have listed above!
c) When the die is fair, this is what happen.
Just for understanding:
Other outcome from the sample space when the die is fair are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
The bold outcome are values when sum are 6 or more. And there are 26 of them.