Math questions

PROBABILITY:-
Q1. A bag contains 6 red, 5 white & 9 black
balls. A ball is drawn at random from bag. The
probability of drawing a red or white ball is-
•11/20
•1/4
•3/4
•7/10
Ans.-11/20

Q2. A dice is thrown once. The probability of
getting even number is
•1/2
•1/3
•1/4
•2/3
Ans.-1/2

Q3. One card is drawn from a well shuffled
deck of cards.The probability of getting a
black
queen is
 1/26
 1/13
 3/52
 26/52
Ans.-1/26

Q4. Two coins are tossed simultaneously.
Find the probability of getting at least 1 tail.
•3/4
•1/4
•1/2
•1
Ans.-3/4

Q5. It is know that a box of 600 ball contains 12 defected
balls. One ball is taken out at random from this box. The
probability of number of deflected balls is-
•49/50
•1/50
•47/50
•1/25
Ans.-49/50

STATISTICS:-
Q1. In the 2nd game of the 1989 World Series between Oakland
and San Francisco 10 players went hitless, 8 players had one hit
apiece, and one player had three hits. What were the mean and
median number of hits?
• 11/19, 0
• 11/19, 1
• 1, 1
• 0, 1
Ans.-11/19,0

Q2. If MEAN=18, tick the correct value of f.
Class
Interval
11-13 13-15 15-17 17-19 19-21 21-23 23-25
fi 7 6 9 13 f 5 4
1.10
2.20
3.30
4.40
Ans.-20

Q3. Suppose the average score on a national test is 500 with a
standard deviation of 100. If each score is increased by 25, what
are the new mean and standard deviation?
•500, 100
•500, 125
•525, 100
•525, 105
•525, 125
Ans.-525,105

Q4. If Q1 = 20 and Q3 = 30, which of the following must be true?
I. The median is 25
II. The mean is between 20 and 30
III. The standard deviation is at most 10.
•I only
•II only
•III only
•All are true
•None are true
Ans.-All are true.

Q5. What is the median of the following
numbers– 83,54,48,60
1. 63.7
2. 63.6
3. 63.5
4. 63.3
Ans:-63.5

 Mean
 Median
 Mode
 Standard deviation
Ans. Standard Deviation

 n+1/2
 n-1/2
 n/2
 n/2+1
Ans. n+1/2

 3
 8
 13
 24
Ans.- 8
Variable x 1 2 3 4 5
Frequency 4 5 y 1 2
Q-If the mean of the following distribution is 2.6, then the
value of y is

 Least frequent value
 Most frequent value
 Middle most value
 None of these
Ans.- Middle Most Value

 A coin is tossed three times.What is the probability that
it lands on heads exactly one time?
 (A) 0.125
(B) 0.250
(C) 0.333
(D) 0.375
(E) 0.500
 The correct answer is (D).
If you toss a coin three times, there are a total of eight possible outcomes.
They are: HHH, HHT, HTH,THH, HTT,THT,TTH, and TTT. Of the eight
possible outcomes, three have exactly one head.They are: HTT,THT, and
TTH.Therefore, the probability that three flips of a coin will produce
exactly one head is 3/8 or 0.375.

Q.Which of the following statements are true? (Check one)
 I. Categorical variables are the same as qualitative variables.
II. Categorical variables are the same as quantitative
variables.
III. Quantitative variables can be continuous variables.
 (A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
 The correct answer is (E).

 Write all the possible outcomes.
 Are the outcomes equally likely?
 find probability of getting a number <
or = 3
 find prob. Of getting a number >6

 When an unbiased dice is thrown
once,the six possible outcomes are:
1,2,3,4,5,6
 Yes, as the dice is unbiased.
 P (a number _<3) = P(1,2,3)=3/6=1/2
 P (a number > 6) = 0/6 =0

 Both heads
 Both heads and both tails
Ans.
1/4
2/4 = 1/2

• Exactly 2 heads
• Head and tail alternatively
Answer:
P (exactly 2 heads)= 3/8
P (head and tail alternatively)= 2/8=1/4

• 3
• 7
• 3 or 7
• Both 3 and 7

 Numbers divisible by 3 are
3,6,9,12,15=5/17
 Numbers divisible by 7 are 7,14=2/17
 Numbers divisible by 3 or 7 are
3,6,9,12,15,7,14=7/17
 There is no number divisible by both 3
and 7.=0/17=0

• Blue ball
• Red ball
• Red or green ball
• Green or blue ball

 Ans.
 5/20=1/4
 10/20=1/2
 10+5/20=3/4
 5+5/20=1/2

 Problem
A student goes to the library.The
probability that she checks out (a) a work
of fiction is 0.40, (b) a work of non-fiction
is 0.30, , and (c) both fiction and non-
fiction is 0.20.What is the probability that
the student checks out a work of fiction,
non-fiction, or both?

 Solution
 Let F = the event that the student checks out fiction; and
let N = the event that the student checks out non-fiction.
When either or both of these events occur, we say that
the union of F and N has occurred.
 This problem requires us to find the probability of the
union of F and N.To find that probability, we use the
rule of additation:
 P(F N) = P(F) + P(N) - P(F N)∪ ∩
P(F N) = 0.40 + 0.30 - 0.20 = 0.50∪

Problem 1
 A card is drawn randomly from a deck of
ordinary playing cards.You win $10 if the
card is a spade or an ace.What is the
probability that you will win the game?

 Solution
 Let S = the event that the card is a spade; and let A = the
event that the card is an ace.You will win the game if either of
these two events occur.Therefore, this problem requires you
to find the probability of the union of two events.
 We know the following:
 There are 52 cards in the deck.
 There are 13 spades, so P(S) = 13/52.
 There are 4 aces, so P(A) = 4/52.
 There is 1 ace that is also a spade, so P(S A) = 1/52.∩
 Therefore, based on the rule of addition:
 P(S A) = P(S) + P(A) - P(S A)∪ ∩
P(S A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13∪

 The correct answer is B.The solution involves four steps.
 Find the mean difference (male absences minus female absences) in the
population.
d = 1 - 2 = 15 - 10 = 5μ μ μ
 Find the standard deviation of the difference.
d = sqrt( 12 / n1 + 22 / n2 ) σ σ σ
d = sqrt(72/100 + 62/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + .72) = sqrt(1.21)σ
= 1.1
 Find the z-score that produced when boys have three more days of absences
than girls.When boys have three more days of absences, the number of male
absences minus female absences is three. And the associated z-score is
z = (x - )/ = (3 - 5)/1.1 = -2/1.1 = -1.818μ σ
 Find the probability.This problem requires us to find the probability that the
average number of absences in the boy sample minus the average number of
absences in the girl sample is less than 3.To find this probability, we enter the
z-score (-1.818) into Stat Trek's Normal Distribution Calculator.We find that
the probability of a z-score being -1.818 or less is about 0.035.
Therefore, the probability that the difference between samples will be no more
than 3 days is 0.035.

 A die is rolled, find the probability that
an even number is obtained.
Let us first write the sample space S of the experiment.
S = {1,2,3,4,5,6}
Let E be the event "an even number is obtained" and write it
down.
E = {2,4,6}
We now use the formula of the classical probability.
P(E) = n(E) / n(S) = 3 / 6 = 1 / 2
Solution to Problem2:

 Two coins are tossed, find the probability
that two heads are obtained.
Note: Each coin has two possible
outcomes H (heads) and T (Tails).

 The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}
 Let E be the event "two heads are obtained".
E = {(H,H)}
 We use the formula of the classical probability.
P(E) = n(E) / n(S) = 1 / 4

 A probability is always greater than or
equal to 0 and less than or equal to 1,
hence only a) and c) above cannot
represent probabilities: -0.00010 is
less than 0 and 1.001 is greater than 1.

Two dice are rolled, find the probability that the sum is
a) equal to 1
b) equal to 4
c) less than 13

 a) The sample space S of two dice is shown below.
S = { (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) }
 Let E be the event "sum equal to 1".There are no outcomes which
correspond to a sum equal to 1, hence
P(E) = n(E) / n(S) = 0 / 36 = 0
 b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),
(3,1)}, hence.
P(E) = n(E) / n(S) = 3 / 36 = 1 / 12
 c) All possible outcomes, E = S, give a sum less than 13, hence.
P(E) = n(E) / n(S) = 36 / 36 = 1

A die is rolled and a coin is tossed, find
the probability that the die shows an
odd number and the coin shows a
head.

 The sample space S of the experiment described in question 5
is as follows
S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)
(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}
 Let E be the event "the die shows an odd number and the coin
shows a head". Event E may be described as follows
E={(1,H),(3,H),(5,H)}
 The probability P(E) is given by
P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

 A card is drawn at random from a deck
of cards. Find the probability of
getting the 3 of diamond.

 The sample space S of the experiment in question 7 is shown below
 Let E be the event "getting the 3 of diamond". An examination of the
sample space shows that there is one "3 of diamond" so that n(E) = 1
and n(S) = 52. Hence the probability of event E occuring is given by
P(E) = 1 / 52

 A card is drawn at random from a deck
of cards. Find the probability of
getting a queen

The sample space S of the
experiment in question 7 is
shwon above (see question 6)
Let E be the event "getting a
Queen". An examination of the
sample space shows that there
are 4 "Queens" so that n(E) = 4
and n(S) = 52. Hence the
probability of event E occuring
is given by
P(E) = 4 / 52 = 1 / 13

 A jar contains 3 red marbles, 7 green
marbles and 10 white marbles. If a
marble is drawn from the jar at
random, what is the probability that
this marble is white?

 We first construct a table of frequencies that gives the marbles color
distributions as follows
 Colorfrequency red3 green7 white10
 We now use the empirical formula of the probability
 Frequency for white color
P(E)=________________________________________________
 Total frequencies in the above table
= 10 / 20 = 1 / 2

 The blood groups of 200 people is
distributed as follows: 50 have
type A blood, 65 have B blood type, 70
have Oblood type and 15 have
type AB blood. If a person from this
group is selected at random, what is
the probability that this person has O
blood type?

 We construct a table of frequencies for the the blood groups as
follows
 Group Frequency a = 50 B = 65 O = 70 AB = 15
 We use the empirical formula of the probability

 Frequency for O blood
 P(E)= ________________________
 Total frequencies

= 70 / 200 = 0.35

 a) A die is rolled, find the probability that
the number obtained is greater than 4.
b) Two coins are tossed, find the probability
that one head only is obtained.
c) Two dice are rolled, find the probability
that the sum is equal to 5.
d) A card is drawn at random from a deck of
cards. Find the probability of getting the
King of heart.

a) 2 / 6 = 1 / 3
b) 2 / 4 = 1 / 2
c) 4 / 36 = 1 / 9
d) 4 / 52 = 1 / 13

 For boys, the average number of absences in the first
grade is 15 with a standard deviation of 7; for girls, the
average number of absences is 10 with a standard
deviation of 6.
In a nationwide survey, suppose 100 boys and 50 girls
are sampled.What is the probability that the male
sample will have at most three more days of absences
than the female sample?
 (A) 0.025
(B) 0.035
(C) 0.045
(D) 0.055
(E) None of the above

 The correct answer is B.The solution involves four steps.
 Find the mean difference (male absences minus female absences)
in the population.
 μd = μ1 - μ2 = 15 - 10 = 5

 Find the standard deviation of the difference.
 σd = sqrt( σ1
2
/ n1 + σ2
2
/ n2 )
σd = sqrt(72
/100 + 62
/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + .72)
= sqrt(1.21) = 1.1

 Find the z-score that produced when boys have three more days of
absences than girls.When boys have three more days of
absences, the number of male absences minus female absences is
three. And the associated z-score is
 z = (x - )/ = (3 - 5)/1.1 = -2/1.1 = -1.818μ σ

Continued…

 Find the probability.This problem requires us to find
the probability that the average number of absences in
the boy sample minus the average number of absences
in the girl sample is less than 3.We find that the
probability of a z-score being -1.818 or less is about
0.035.
 Therefore, the probability that the difference between
samples will be no more than 3 days is 0.035.
Continued…

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