1. An urn B1 contains 2 white and 3 black chips and another urn B2 contains 3 white and 4 black chips. One urn is selected at random and a chip is drawn from it. If the chip drawn is found black, find the probability that the urn chosen was B1.
a) 4⁄7
b) 3⁄7
c) 20⁄41
d) 21⁄41
Explanation: Let E1, E2 denote the vents of selecting urns B1 and B2 respectively.
Then P(E1) = P(E2) = 1⁄2
Let B denote the event that the chip chosen from the selected urn is black .
Then we have to find P(E1 /B).
By hypothesis P(B /E1) = 3⁄5
and P(B /E2) = 4⁄7
By Bayes theorem P(E1 /B) = (P(E1)*P(B│E1))/((P(E1) * P(B│E1)+P(E2) * P(B│E2)) )
= ((1/2) * (3/5))/((1/2) * (3/5)+(1/2)*(4/7) ) = 21/41.
2. At a certain university, 4% of men are over 6 feet tall and 1% of women are over 6 feet tall. The total student population is divided in the ratio 3:2 in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?
a) 2⁄5
b) 3⁄5
c) 3⁄11
d) 1⁄100
Explanation: Let M be the event that student is male and F be the event that the student is female. Let T be the event that student is taller than 6 ft.
P(M) = 2⁄5 P(F) = 3⁄5 P(T|M) = 4⁄100 P(T|F) = 1⁄100
P(F│T) = (P(T│F) * P(F))/(P(T│F) * P(F) + P(T│M) * P(M))
= ((1/100) * (3/5))/((1/100) * (3/5) + (4/100) * (2/5) )
= 3⁄11.
3. Previous probabilities in Bayes Theorem that are changed with help of new available information are classified as _________________
a) independent probabilities
b) posterior probabilities
c) interior probabilities
d) dependent probabilities
Explanation: posterior probabilities
4. Consider a dice with the property that that probability of a face with n dots showing up is proportional to n. The probability of face showing 4 dots is?
a) \(\frac{1}{7} \)
b) \(\frac{5}{42} \)
c) \(\frac{1}{21} \)
d) \(\frac{4}{21} \)
Explanation: P (n) is proportional to n where n=
1,2,3,…6 is random variable.
P(n) = kn
P(1)+P(2)….P (6) = 1
K(1+2+3+4+5+6) = 1
\(K=\frac{1}{21} \)
Hence P(4) = 4K = \(\frac{4}{21}. \)
5. Let X be a random variable with probability distribution function f (x)=0.2 for |x|<1
= 0.1 for 1 < |x| < 4
= 0 otherwise
The probability P (0.5 < x < 5) is _____
a) 0.3
b) 0.5
c) 0.4
d) 0.8
Explanation: P (0.5 < x < 5) = Integrating f (x) from
0.5 to 5 by splitting in 3 parts that is from 0.5 to 1
and from 1 to 4 and 4 to 5 we get
P (0.5 < x < 5) = 0.1 + 0.3 + 0
P (0.5 < x < 5) = 0.4.
6. Runs scored by batsman in 5 one day matches are 50, 70, 82, 93, and 20. The standard deviation is ______
a) 25.79
b) 25.49
c) 25.29
d) 25.69
Explanation: The mean of 5 innings is
(50+70+82+93+20)÷5 = 63
S.D = [1⁄n (x(n)-mean)2]0.5
S.D = 25.79.
7. Find median and mode of the messages received on 9 consecutive days 15, 11, 9, 5, 18, 4, 15, 13, 17.
a) 13, 6
b) 13, 18
c) 18, 15
d) 15, 16
Explanation: Arranging the terms in ascending order 4, 5, 9, 11, 13, 14, 15, 18, 18.
Median is \(\frac{(n+1)}{2} \) term as n = 9 (odd) = 13.
Mode = 18 which is repeated twice.
8. Mode is the value of x where f(x) is a maximum if X is continuous.
a) True
b) False
Explanation: For a continuous variable mode is defined as the value where f(x) is a maximum or it is defined as the quantity repeated maximum number of times.
9. E (XY)=E (X)E (Y) if x and y are independent.
a) True
b) False
Explanation: By the property of Expectation
E (XY) = E (X) E (Y).
That is the Expectation of a composite function XY is the product of the individual expectations of X and Y.
10. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is ______
a) \(\frac{1}{2} \)
b) \(\frac{1}{3} \)
c) \(\frac{1}{4} \)
d) \(\frac{1}{6} \)
Explanation: p=0.5 (Probability of tail)
q=1-0.5=0.5
n=4 and x is binomial variate.
P (X=x) = nCx px qn-x.
P (X=3) = 4C3 (0.5)3 = 1⁄2.