Probability MCQ Questions & Answers in Statistics and Probability | Maths
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101.
Three numbers are chosen at random without replacement from the set $$A = \left\{ {x|1 \leqslant x \leqslant 10,\,x \in N} \right\}.$$ The probability that the minimum of the chosen numbers is $$3$$ and maximum is $$7$$, is :
A
$$\frac{1}{{12}}$$
B
$$\frac{1}{{15}}$$
C
$$\frac{1}{{40}}$$
D
none of these
Answer :
$$\frac{1}{{40}}$$
$$n\left( S \right) = {}^{10}{C_3}$$ and $$n\left( E \right) = {}^3{C_1},$$ because on selecting $$3,\,7$$ and we have to select one from $$4,\,5$$ and $$6.$$
$$\therefore \,P\left( E \right) = \frac{{{}^3{C_1}}}{{{}^{10}{C_3}}} = \frac{1}{{40}}.$$
102.
If the random variable $$X$$ takes the values $${x_1},\,{x_2},\,{x_3},.....,\,{x_{10}}$$ with probabilities $$P\left( {X = {x_i}} \right) = k\,i,$$ then the value of $$k$$ is equal to :
103.
The probabilities of two events $$A$$ and $$B$$ are given as $$P\left( A \right) = 0.8$$ and $$P\left( B \right) = 0.7.$$ What is the minimum value of $$P\left( {A \cap B} \right)\,?$$
104.
Three persons $$A,\,B$$ and $$C$$ are to speak at a function along with five others. If they all speak in random order, the probability that $$A$$ speaks
before $$B$$ and $$B$$ speaks before $$C$$, is :
A
$$\frac{3}{8}$$
B
$$\frac{1}{6}$$
C
$$\frac{3}{5}$$
D
None of these
Answer :
$$\frac{1}{6}$$
The total number of ways in which $$8$$ persons can speak is $${}^8{P_8} = 8!.$$
The number of ways in which $$A,\,B$$ and $$C$$ can be arranged in the specified speaking order is $${}^8{C_3}.$$
There are $$5!$$ ways in which the other five can speak.
So, favourable number of ways is $${}^8{C_3} \times 5!.$$
Hence, required probability $$ = \frac{{{}^8{C_3} \times 5!.}}{{8!}} = \frac{1}{6}.$$
105.
A bag contains $$14$$ balls of two colours, the number of balls of each colour being the same. $$7$$ balls are drawn at random one by one. The ball in hand is returned to the bag before each new draw. If the probability that at least $$3$$ balls of each colour are drawn is $$p$$ then :
A
$$p > \frac{1}{2}$$
B
$$p = \frac{1}{2}$$
C
$$p < 1$$
D
$$p < \frac{1}{2}$$
Answer :
$$p > \frac{1}{2}$$
A ball drawn is of either colour. If drawing a ball of one colour is a success then drawing ball of the other colour is a failure.
$$\therefore $$ the probability of success in one draw $$ = \frac{7}{{14}} = \frac{1}{2}$$ and the probability of failure in one draw $$ = \frac{7}{{14}} = \frac{1}{2}.$$
$$\therefore $$ the probability that at least $$3$$ balls of each colour is drawn
$$\eqalign{
& = {}^7{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^4} + {}^7{C_4}.{\left( {\frac{1}{2}} \right)^4}.{\left( {\frac{1}{2}} \right)^3} \cr
& = 2.\frac{{7!}}{{\left( {3!} \right)\left( {4!} \right)}}.\frac{1}{{{2^7}}} \cr
& = 2.\frac{{7.6.5}}{6}.\frac{1}{{{2^7}}} \cr
& = \frac{{35}}{{64}} > \frac{1}{2} \cr} $$
106.
A man draws a card from a pack of $$52$$ cards and then replaces it. After
shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw is :
A
$$\frac{9}{{64}}$$
B
$$\frac{{27}}{{64}}$$
C
$$\frac{1}{4} \times \frac{{{}^{39}{C_2}}}{{{}^{52}{C_2}}}$$
D
none of these
Answer :
$$\frac{9}{{64}}$$
Let $$E = $$ the event of drawing a heart. Clearly, $$P\left( E \right) = \frac{{{}^{13}{C_1}}}{{{}^{52}{C_1}}} = \frac{1}{4}.$$
$$\therefore $$ the required probability $$ = P\left( {\overline E \,\overline E \,E} \right) = P\left( {\overline E } \right).P\left( {\overline E } \right).P\left( E \right) = \frac{3}{4}.\frac{3}{4}.\frac{1}{4} = \frac{9}{{64}}.$$
107.
An experiment has 10 equally likely outcomes. Let $$A$$ and $$B$$ be non-empty events of the experiment. If $$A$$ consists of 4 outcomes, the number of outcomes that $$B$$ must have so that $$A$$ and $$B$$ are independent, is
A
2, 4 or 8
B
3, 6 or 9
C
4 or 8
D
5 or 10
Answer :
5 or 10
We have $$n\left( S \right) = 10,n\left( A \right) = 4$$
Let $$n\left( B \right) = x\,\,{\text{and }}n\left( {A \cap B} \right) = y$$
Then for $$A$$ and $$B$$ to be independent events
$$\eqalign{
& P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right) \cr
& \Rightarrow \,\,\frac{y}{{10}} = \frac{4}{{10}} \times \frac{x}{{10}} \cr
& \Rightarrow \,\,x = \frac{5}{2}y \cr} $$
⇒ $$y$$ can be 2 or 4 so that $$x$$ = 5 or 10
$$\therefore \,\,n\left( B \right) = 5\,\,{\text{or 10}}$$
108.
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
109.
Consider a set $$P$$ containing $$n$$ elements. A subset $$A$$ of $$P$$ is drawn and there after set $$P$$ is reconstructed. Now one more subset $$B$$ of $$P$$ is drawn. Probability of drawing sets $$A$$ and $$B$$ so that $$A \cap B$$ has exactly one element is :
A
$${\left( {\frac{3}{4}} \right)^n}.n$$
B
$$n.{\left( {\frac{3}{4}} \right)^{n - 1}}$$
C
$$\left( {n - 1} \right).{\left( {\frac{3}{4}} \right)^n}$$
Let $${x_i}$$ be any element of set $$P$$, we have following possibilities
$$\eqalign{
& \left( {\bf{i}} \right)\,\,{x_i}\, \in \,A,\,{x_i}\, \in \,B\,; \cr
& \left( {{\bf{ii}}} \right)\,\,{x_i}\, \in \,A,\,{x_i}\, \notin \,B\,; \cr
& \left( {{\bf{iii}}} \right)\,\,{x_i}\, \notin \,A,\,{x_i}\, \in \,B\,; \cr
& \left( {{\bf{iv}}} \right)\,\,{x_i}\, \notin \,A,\,{x_i}\, \notin \,B \cr} $$
Clearly, the element $${x_i}\, \in \,A \cap B$$ if it belongs to $$A$$ and $$B$$ both. Thus out of these $$4$$ ways only first way is favourable. Now the element that we want to be in the intersection can be chosen in $$'n'$$ different ways.
Hence required probability is $$n.{\left( {\frac{3}{4}} \right)^{n - 1}}.$$
110.
If two different numbers are taken from the set (0, 1,2, 3, . . . . . , 10), then the probability that their sum as well as absolute difference are both multiple of 4, is:
A
$$\frac{{7}}{{55}}$$
B
$$\frac{{6}}{{55}}$$
C
$$\frac{{12}}{{55}}$$
D
$$\frac{{14}}{{55}}$$
Answer :
$$\frac{{6}}{{55}}$$
Let $$A$$ $$ \equiv $$ {0, 1, 2, 3, 4, . . . . . , 10}
$$n\left( S \right) = {\,^{11}}{C_2} = 55$$ where $$'S'$$ denotes sample space
Let $$E$$ be the given event
∴ $$E$$ $$ \equiv $$ {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
⇒ $$n(E) = 6$$
$$\therefore \,\,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{6}{{55}}$$