Glen Davis is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one,
2 answers:
Answer:
0.00457 or 0.457%
Step-by-step explanation:
Glen Davis makes a free throw 74% of the time, meaning he misses 26% of the time. Four him to miss four in a row, it would require him to miss his first throw and his second throw and his third throw and his fourth throw. I highlight and because it is a helpful tip to remember that AND situations require multiplication while OR situations require addition.
So in this question we multiply the probabilities of him missing.
This answer has been rounded off to three significant figures
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Answer:
She sells 4 loaves of bread
Step-by-step explanation:
Lets explain how to solve the problem
→ A baker uses cups of flour to make bread
→ She uses cups of flour to make each loaf
From these information we can find the number of loaves of bread
she can make
∵ There are cups of flour
∵ Each loaf of bread needs cups of flour
∴ The number of loaves = ÷
To divide two mixed numbers make them improper fractions and
change the division sign to multiplication sign and reciprocal the
fraction after the division sign
∵ =
∴ =
∵ =
∴ =
∴ The number of loaves = ×
∴ The number of loaves = 6
<em>She can make 6 loaves</em>
→ The baker sells of the loaves of bread that she makes
→ We need to find the number of loves of bread she sells
∵ She sells of the loaves
∵ There are 6 loves
∴ The number of loaves she sells = 6 × = 4
<em>She sells 4 loaves of bread</em>
Answer:
The probability that of the two chips selected both are defective is 0.1089.
Step-by-step explanation:
Let <em>X</em> = number of defective chips.
It is provided that there are 2 defective chips among 6 chips.
The probability of selecting a defective chip is:
A sample of <em>n</em> = 2 chips are selected.
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 2 and <em>p</em> = 0.33.
The probability function of a Binomial distribution is:
Compute the probability that of the two chips selected both are defective as follows:
Thus, the probability that of the two chips selected both are defective is 0.1089.
The sample space of selecting two chips is:
S = (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)