1) The traffic data of New York City on New Year's Eve shows that the probability of a person having a car accident is 0.07. The probability of a person driving while intoxicated is 0.49 and probability of a person having a car accident while intoxicated is 0.18.
a) What is the probability of a person driving while intoxicated or having a car accident?
b) Which rule is applied here and why?
2). In a class of 50 students, 5 students got A grade, 22 got B grade, 15 got C grade and 8 got D grade. Find the following probabilities.
i). Neither A nor D
ii). Not B
iii). A or C
I'm in the middle of my stats paper. Kindly solve these questions with explanation.?
2020-08-20 3:46 pm
回答 (2)
2020-08-20 4:14 pm
✔ 最佳答案
1)a)
A: a person driving while intoxicated, P(A) = 0.49
B: having a car accident, P(B) = 0.07
P(A and B) = 0.18
P(A or B)
= P(A) + P(B) - P(A and B)
= 0.49 + 0.07 - 0.18
= 0.38
b)
The rule is: P(A and B) = P(A) + P(B) - P(A or B)
This is because the two events (A and B) are not mutually exclusive.
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2)
n(total) = 50
n(A) = 5
n(B) = 22
n(C) = 15
n(D) = 8
i)
n(Neither A nor D)
= n(total) - [n(A) + n(D)]
= 50 - (5 + 8)
= 37
ii)
n(Not B)
= n(total) - n(B)
= 50 - 22
= 28
iii)
n(A or C)
= n(A) + n(C)
= 5 + 15
= 20
2020-08-20 4:52 pm
b) Rule => P(I or A) = P(I) + P(A) - P(I and A)
a) so, 0.49 + 0.07 - 0.18 => 0.38
A or D = 13/50
so, neither A nor D = 37/50
P(B) = 22/50
so, P(not B) = 28/50 = 14/25
P(A or C) = (5 + 15)/50 = 20/50 = 2/5
:)>
a) so, 0.49 + 0.07 - 0.18 => 0.38
A or D = 13/50
so, neither A nor D = 37/50
P(B) = 22/50
so, P(not B) = 28/50 = 14/25
P(A or C) = (5 + 15)/50 = 20/50 = 2/5
:)>
收錄日期: 2021-04-24 08:01:15
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