1. The probability that a student has a Visa card (event V) is 0.30. The probabi
ID: 3369599 • Letter: 1
Question
1. The probability that a student has a Visa card (event V) is 0.30. The probability that a student has a MasterCard (event M) is 0.40. The probability that a student has both cards is 0.12. (1) Find the probability that a student has either a Visa card or a MasterCard. (2) In this problem, are V and M independent? Why?
2.
This is a contingency table describes 100 business students.
Gender
Major
Female(F)
Male(M)
Accounting (A)
22
28
Economics(E)
16
14
Statistics(S)
12
8
Find each probability.
(1) P(A)
(2) P(A?M)
(3) P(A|F)
(4) P(F|S)
(5) P(E?F)
3.
Suppose that 12% of the women who purchase over-the-counter pregnancy testing kits are actually pregnant. For a particular brand of kit, if a woman is pregnant, the test will yield a positive result 96% of the time and a negative result 4% of the time. If she is not pregnant, the test will yield a positive result 5% of the time and a negative result 95% of the time.
Suppose the test comes up positive.
Let event A be Positive test.
Let event B be Pregnant.
(1) P(A|B) =?
(2) P(A|B')=?
(3) What is the probability that she is really pregnant?
(i.e. What is P(B|A) (the probability that she is really pregnant under the condition that the test comes up positive)?)
4.
Consider a random experiment “tossing un-fair coin two times” where the probability of head for the unfair coin is 0.4. Let random variable X be the number of heads. That is, X follows the binomial distribution with parameters n = 2, ?=0.4.
(1) Calculate P(X?1).
(2) Calculate P(X>1).
(3) Find the expected value of X.
(4) Find the standard deviation of X.
Gender
Major
Female(F)
Male(M)
Accounting (A)
22
28
Economics(E)
16
14
Statistics(S)
12
8
Explanation / Answer
Ans:
1)
Given that
P(V)=0.3
P(M)=0.4
P(V and M)=0.12
a)P(V or M)=P(V)+P(M)-P(V and M)
=0.3+0.4-0.12
=0.58
b)P(V and M)=0.12
P(V)*P(M)=0.3*0.4=0.12
As,P(V and M)=P(V)*P(M),V and M are independent.
2)
i)P(A)=(22+28)/100=0.5
ii)P(A and M)=28/100=0.28
iii)P(A/F)=P(A and F)/P(F)=22/(22+16+12)=0.44
IV)P(F/S)=P(F and S)/P(S)=12/(12+8)=0.6
V)P(E or F)=P(E)+P(F)-P(E and F)
P(E)=(16+14)/100=0.3
P(F)=(22+16+12)/100=0.5
P(E and F)=16/100=0.16
So,
P(E or F)=0.3+0.5-0.16=0.64
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.