Probability Test

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Probability Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Roll a fair die twice. Let $$A$$ be the even that the sum of the two rolls equals six, and let $$B$$ be the even that the same number comes up twice. What is $$P(A/B)$$?

A
$$\dfrac16$$

B
$$\dfrac56$$

C
$$\dfrac15$$

D
none of these

Solution

Favorable out comes for event $$A=\left(1,5\right) \left(4,2\right) \left(3,3\right) \left(4,2\right) \left(5,1\right)$$

Favorable out comes for event $$B=\left(1,1\right) \left(2,2\right) \left(3,3\right) \left(4,4\right) \left(5,5\right) \left(6,6\right)$$

Favorable out comes for event $$A\cap B=\left(3,3\right)$$

$$P\left(\dfrac{A}{B}\right)=\dfrac{n\left(A\cap B\right)}{n\left(B\right)}=\dfrac{1}{6}$$

$$\therefore$$ option $$A$$ is correct.
Question 2
1 / -0

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is :

A
$$\dfrac{10}{3^5}$$

B
$$\dfrac{17}{3^5}$$

C
$$\dfrac{13}{3^5}$$

D
$$\dfrac{11}{3^5}$$

Solution

$${\textbf{Step - 1: Simplify the problems to solve using probability}}$$
$${\text{We know that we have 3 alternative answer but only 1 is correct }}$$
$$\therefore {\text{ The probability of getting a correct answer }}\dfrac{1}{3}$$
$${\text{So the probability of getting an incorrect answer is }}\left( {1 - \dfrac{1}{3}} \right)$$$$ = \dfrac{2}{3}$$
$${\textbf{Step - 2: Use binomial distribution }}$$
$${\text{So the probability of getting 4 or correct answer = }}$$
$${\text{probability of 4 or more correct + probability of 5 correct answer }}..........{\text{(i)}}$$
$${\text{Now we can say , for probability of 4 correct answer x = 4 \& n = 5}}$$
$${\text{By binomial distribution , }}$$
$${\text{Probability of getting 4 correct answer }} = {}^5{C_4}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{5 - 4}}$$
$$ = 5 \times {\left( {\dfrac{1}{3}} \right)^4} \times \left( {\dfrac{2}{3}} \right)$$
$$ = 5 \times \dfrac{2}{{{3^5}}}$$
$${\text{For probability of 5 correct answer x}} = 5{\text{ \& n}} = 5$$
$${\text{So the probability of getting 5 correct answers }} = {}^5{C_5} \times {\left( {\dfrac{1}{3}} \right)^5} \times {\left( {\dfrac{2}{3}} \right)^{5 - 5}}$$
$${\text{Probability of getting 5 correct answers }} = {\left( {\dfrac{1}{3}} \right)^5}$$
$${\textbf{Step - 3: Put the values in (i)}}$$
$${\text{So the probability of getting 4 or more correct answers }} = 5 \times \dfrac{2}{{{3^5}}} + \dfrac{1}{{{3^5}}}$$
$$ = \dfrac{{10}}{{{3^5}}} + \dfrac{1}{{{3^5}}}$$
$$ = \dfrac{{11}}{{{3^5}}}$$
$${\textbf{Hence, the probability that a student will get 4 or more answer correct answer is }}\mathbf{\dfrac{{11}}{{{3^5}}}}$$
Question 3
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If two events A and B are such that $$P ( \overline { A } ) = \dfrac { 3 } { 10 } , P ( B ) = \dfrac { 2 } { 5 }$$ and $$P ( A \cap \overline { B } ) = \dfrac { 1 } { 2 }$$ ,then $$P \left( \dfrac { B } { A \cup \overline { B } } \right)$$ is

A
$$\dfrac { 1 } { 2 }$$

B
$$\dfrac { 1 } { 3 }$$

C
$$\dfrac { 1 } { 4 }$$

D
$$\dfrac { 2 } { 3 }$$

Solution

$$ P(\bar{A}) = 3/10 \Rightarrow P(A) = 7/10 $$

$$ P(B) = 2/5 \Rightarrow P(B) = 3/5 $$

$$ P(A\cup B) = P(\bar{A})+P(\bar{B})-P(A\cap \bar{B}) $$

$$ = 7/10+3/5-1/2 $$

$$ = 4/5 $$

$$P(B/(A\cup B)) = \dfrac{P(B\cap (A\cup \bar{B}))}{P(A\cup B)} $$

$$ P(B\cap (A\cup \bar{B})) = P(B)+P(A\cup B)-P(B\cup (A\cup \bar{B})) $$

$$ P(B\cup \bar{B}) = 1 \Rightarrow P(B\cup A\cup \bar{B}) = 1 $$

$$ \therefore P(B\cap (A\cup B)) = 2/5+4/5-1 $$

$$ = 1/5 $$

$$ \therefore P(B/(A\cup \bar{B})) = \frac{1/5}{4/5} = 1/4 $$
Question 4
1 / -0

A natural number is chosen at random from amongest the first $$500$$. The probability that the number chosen is divisible by $$3$$ or $$5$$ is ? .

A
$$\dfrac{1}{5}$$

B
$$\dfrac{{33}}{{500}}$$

C
$$\dfrac{{83}}{{250}}$$

D
$$\dfrac{{233}}{{500}}$$

Solution

Multiples of $$3$$ among first $$500$$ number$$=\{3, 6, 9,......498\}$$
$$n(3)=166$$
Multiples of $$5$$ among first $$500$$ numbers$$=\{5, 10, 15,...….500\}$$
$$n(5)=100$$
Multiples of $$3$$ and $$5=\{15, 30, 45, 495\}$$
$$n(345)=33$$
$$n(3$$ or $$5)=n(3)+n(5)-n(3$$ & $$5)$$
$$=166+100-33=233$$
$$\therefore$$ Probability that number is divisible by $$3$$ or $$5=\dfrac{233}{500}$$.
Question 5
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If $$C\ and D$$ are two events such that $$C\subset D$$ and $$P\left( D \right) \neq 0$$, then the correct statement among the following is:

A
$$P\left (C/D\right)=P\left (C\right)$$

B
$$ P\left( C/D \right) \ge P\left( C \right)$$

C
$$P(C/D) \leq P(C)$$

D
$$ P\left( C/D \right) =\dfrac { P\left( D \right) }{ P\left( C \right) }$$

Solution

$$ P\left( C|D \right)=\frac{P\left( C\cap D \right)}{P\left( D \right)} $$
$$ as\,C\subset D,P\left( C \right)\subset P\left( D \right) $$
$$ \therefore P\left( C\cap D \right)=P\left( C \right) $$
$$ We\,have\,P\left( C|D \right)=\frac{P\left( C \right)}{P\left( D \right)} $$
$$ As0<P\left( D \right)\le 1 $$
$$ Hence $$
$$ P\left( C|D \right)\ge P\left( C \right) $$
Question 6
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A letter is known to have come either from $$'KRISHNAGIRI'$$ or $$'DHARMPURI'$$. On the post mark only the two consecutive letters $$'RI$$ are visible. Then, the chance that it come from $$'KRISHNAGIRI'$$ is

A
$$\dfrac {3}{5}$$

B
$$\dfrac {2}{3}$$

C
$$\dfrac {9}{14}$$

D
$$\dfrac {5}{14}$$

Solution
Question 7
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A random variable $$X$$ has the probability distribution:
$$X$$: 1 2 3 4 5 6 7 8
$$p(X)$$: 0.2 0.2 0.1 0.1 0.2 0.1 0.1 0.1
For the events $$E=\left\{ X\quad is\quad a\quad prime\quad number \right\} $$ and $$F=\left\{ X<4 \right\} $$, the $$P(E\cup F)$$ is

A
$$0.50$$

B
$$0.77$$

C
$$0.35$$

D
$$0.80$$

Solution

given I = $$\{$$'x' is a prime number $$\}$$
P(E) = p(x=2 or x = 3 or x =5 or x =7)
$$=P(x=2)+p(x=3)+p(x=5)+p(x=7)$$
$$ = 0.2+0.1 +0.2+0.1$$
$$ p(B) = 0.6$$
$$ p(F) = P(x = 1)+p(x=2)+p(x =3)$$
$$ = 0.2+0.2+0.1$$
$$ = 0.5$$
$$ p(EAF) = p(x=2)+p(x=3)$$
$$ = 0.2+0.1$$
$$ 0.3 $$
$$p(E\vee F)=p(E)+p(F)-p(E\wedge F)$$
$$ = 0.6+0.5 = 0.3$$
$$ P(E\vee F) = 0.80 $$
Question 8
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If $$A$$ and $$B$$ are two events such that $$P(A)=0.3,\ P(B)=0.6$$ and $$P(B/A)=0.5$$ then $$P(A/B)$$=

A
$$\dfrac 12$$

B
$$\dfrac 13$$

C
$$\dfrac 14$$

D
$$\dfrac 23$$

Solution

$$P(A)=0.3\quad P(B)=0.6$$

$$ \\\\P(B/A)=0.5=\dfrac { P(A\cap B) }{ P(A) } $$

$$ P(A\cap B)=0.5\times P(A)=0.15$$

$$\displaystyle P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.15 }{ 0.6 } =0.25=\frac { 1 }{ 4 } $$
Question 9
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The total number of ways of selecting six coins out of 20 one-rupee coins, 10 fifty-paise coins, and 7 twenty- five paise coins is:

A
28

B
56

C
$$^{37}C_{6}$$

D
none of these

Solution

Using multinomial theorem, are have
$$ x_{1}+x_{2}+x_{3} = 6$$ where $$0\leq x_{1}\leq 6\, 0\leq x_{2} \leq 6\, 0\leq x_{3}\leq 0.$$
$$ \Rightarrow $$ Total no. of ways is $$ ^{n+r-1}C_{r}$$
$$ = ^{3+6-1}C_{6} = ^{8}C_{6}$$
$$ = 28 $$
Question 10
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A natural number $$x$$ is chosen at random from the first $$120$$ natural numbers and it is observed to be divisible by $$8 ,$$ then the probability that it is not divisible by $$6$$ is

A
$$\frac { 1 } { 3 }$$

B
$$\frac { 1 } { 4 }$$

C
$$\frac { 3 } { 4 }$$

D
$$\frac { 2 } { 3 }$$

Solution

Numbers which are divisible by both $$6$$ and $$8$$ are $$24,48,72,96,120$$
So, $$P$$ (Divisible by $$8$$ not by $$6$$) $$=1-\cfrac{Prob(divisible \text{ }by \text{ }both)}{Prob(divisible \text{ }by\text{ }8)}$$
Numbers divisible by $$8$$ are $$8,16\dots, 120$$
So, $$120=8+(n-1)8n=15$$
So, $$P$$ (Divisible by $$8$$ no by $$6$$)$$=1-\cfrac{\cfrac{5}{120}}{\cfrac{15}{120}}=1-\cfrac{1}{3}=\cfrac{2}{3}$$

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$$X$$:	1	2	3	4	5	6	7	8
$$p(X)$$:	0.2	0.2	0.1	0.1	0.2	0.1	0.1	0.1

Probability Test - 39

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Probability Test - 39

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SHARING IS CARING

Question 1

$$\dfrac16$$

$$\dfrac56$$

$$\dfrac15$$

none of these

Solution

Question 2

$$\dfrac{10}{3^5}$$

$$\dfrac{17}{3^5}$$

$$\dfrac{13}{3^5}$$

$$\dfrac{11}{3^5}$$

Solution

Question 3

$$\dfrac { 1 } { 2 }$$

$$\dfrac { 1 } { 3 }$$

$$\dfrac { 1 } { 4 }$$

$$\dfrac { 2 } { 3 }$$

Solution

Question 4

$$\dfrac{1}{5}$$

$$\dfrac{{33}}{{500}}$$

$$\dfrac{{83}}{{250}}$$

$$\dfrac{{233}}{{500}}$$

Solution

Question 5

$$P\left (C/D\right)=P\left (C\right)$$

$$ P\left( C/D \right) \ge P\left( C \right)$$

$$P(C/D) \leq P(C)$$

$$ P\left( C/D \right) =\dfrac { P\left( D \right) }{ P\left( C \right) }$$

Solution

Question 6

$$\dfrac {3}{5}$$

$$\dfrac {2}{3}$$

$$\dfrac {9}{14}$$

$$\dfrac {5}{14}$$

Solution

Question 7

$$0.50$$

$$0.77$$

$$0.35$$

$$0.80$$

Solution

Question 8

$$\dfrac 12$$

$$\dfrac 13$$

$$\dfrac 14$$

$$\dfrac 23$$

Solution

Question 9

28

56

$$^{37}C_{6}$$

none of these

Solution

Question 10

$$\frac { 1 } { 3 }$$

$$\frac { 1 } { 4 }$$

$$\frac { 3 } { 4 }$$

$$\frac { 2 } { 3 }$$

Solution

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