Probability Test

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Probability Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

There is a probability that 4 out of 10 students appearing in CA CPT exams will qualify for CA course after CA CPT exam. The probability that at least of the five students will join CA course is.......

A
$$0.92$$

B
$$0.66$$

C
$$0.45$$

D
$$0.08$$

Solution

Probability that the students will qualify, $$p=\dfrac{4}{10}$$
Probability that they can't qualify, $$p=\dfrac{6}{10}$$
$$\Rightarrow n=5$$
By Bernoulli's equation,
$$\Rightarrow p={^nC_rp^rq^{n-r}}$$
p ( at least one will get selected )
$$={^5C_1}\left( \dfrac { 4 }{ 10 } \right) ^{ 1}\left( \dfrac { 6 }{ 10 } \right) ^{ 4}+{^5C_2}\left( \dfrac { 4 }{ 10 } \right) ^{ 2}\left( \dfrac { 6 }{ 10 } \right) ^{ 3}+{^5C_3}\left( \dfrac { 4 }{ 10 } \right) ^{ 3}\left( \dfrac { 6 }{ 10 } \right) ^{ 2}+{^5C_4}\left( \dfrac { 4 }{ 10 } \right) ^{ 4}\left( \dfrac { 6 }{ 10 } \right)+{^5C_5}\left( \dfrac { 4 }{ 10 } \right) ^{ 5}$$
$$=\dfrac{5\times4\times1296+10\times16\times216+10\times64\times36+5\times256\times6+1024}{100000}$$
$$=0.922.$$
Hence, the answer is $$0.92.$$
Question 2
1 / -0

The range of a random variable X = $${ 1,2,3,4,...}$$ and the probabilities are given by $$ P (X = k) = \frac{C^k}{ k! } $$ ; $$k = 1,2,3,4...,$$ then the value of C is

A
$$2$$

B
$$log_2 e$$

C
$$log_e 2$$

D
4

Solution
Question 3
1 / -0

The probability that Sania wims Wimbledon tournaments final is $$\dfrac{1}{3}$$. If Sania Mirza plays $$3$$ round of Wimbledon final, the probability that she wins at least one round of match is:

A
$$\dfrac{8}{27}$$

B
$$\dfrac{19}{27}$$

C
$$\dfrac{10}{27}$$

D
$$\dfrac{17}{27}$$

Solution

Probability that Sania wins Wins Wimbledon tournaments final $$=\dfrac{1}{2}$$
$$\Rightarrow$$ Probability that Sania losses $$=\dfrac{2}{3}$$
$$\Rightarrow$$ i.e, $$p=\dfrac{1}{3}$$ $$q=\dfrac{2}{3}$$
$$\Rightarrow$$ Total number of rounds, $$n=3$$
$$\Rightarrow$$ By Bernoulli's theorm, $$P={^nC_r}{p^r}q^{n-r}$$
P (She wins atleast $$1$$ round )
$$={^3C_1}\left( \dfrac { 1 }{ 3 } \right) ^{ 1 }\left( \dfrac { 2 }{ 3 } \right) ^{ 2 }+{^3C_2}\left( \dfrac { 1 }{ 3 } \right) ^{ 2 }\left( \dfrac { 2 }{ 3 } \right)+{^3C_3}\left( \dfrac { 1 }{ 3 } \right) ^{ 3 }\left( \dfrac { 2 }{ 3 } \right) ^{ 0 }$$
$$=\dfrac{3\times4}{27}+\dfrac{3\times2}{27}+\dfrac{1}{27}$$
$$=\dfrac{12+6+1}{27}$$
$$=\dfrac{19}{27}$$
Hence, the answer is $$\dfrac{19}{27}.$$
Question 4
1 / -0

The probability of an even happens in one trial of an experiment is $$0.3$$. Three independent trials of the experiments are performed. Find the probability that the event A happens at least once.

A
$$0.657$$

B
$$0.965$$

C
$$0.796$$

D
$$0.509$$

Solution

Probability of an event happens in $$1$$ trial $$=0.3$$
$$\Rightarrow$$ Probability of an event that don't happen in $$1$$ trail $$=0.7$$
$$\Rightarrow$$ Total number of trial , $$n=3$$
$$\Rightarrow$$ $$p=0.3$$ $$q=0.7$$
$$\Rightarrow$$ By Bernoulli's theorm, $$P={^nC_r}{p^r}q^{n-r}$$
P ( Event happens alteast once )
$$={^3C_1}\left( 0.3 \right) ^{ 1 }\left( 0.7 \right) ^{ 2 }+{^3C_2}\left( 0.3 \right) ^{ 2 }\left( 0.7 \right)+{^3C_3}\left( 0.3 \right) ^{ 3 }\left( 0.7 \right) ^{ 0 }$$
$$=0.441+0.189+0.027$$
$$=0.657$$
Hence, the answer is $$0.657.$$
Question 5
1 / -0

If the range of random variable X is {0, 1, 2, 3, 4...} with p(X = k) = $$\frac{(k + 1)a}{3^k}$$ for $$k>=$$0, then a =

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

B
$$\frac{4}{9}$$

C
$$\frac{8}{27}$$

D
$$\frac{16}{81}$$

Solution
Question 6
1 / -0

There is a three-volume dictionary among 43 books arranged on a shelf in random order. Three books are drawn at random from the shelf. The probability that all the three volumes of the dictionary will be drawn, is

A
3/12341

B
2/12341

C
1/12341

D
None of these
Question 7
1 / -0

A pair of dice is rolled together till a sum of either $$5$$ or $$7$$ is obtained. Find the probability that $$5$$ comes before $$7$$.

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

B
$$\frac{1}{5}$$

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

D
$$\frac{2}{5}$$

Solution

$${\textbf{Step - 1: Find probabilities of getting a sum of 5 or 7}}{\text{.}}$$
$${\text{Let }}A{\text{ be the event that the sum of numbers rolled over dice is 5,}}$$
$${\text{The favourable cases will be (1,4),(4,1),(2,3),(3,2) and total number of cases = 6}} \times {\text{6 = 36}}$$
$${\therefore\text{ } } P(A) = \dfrac{4}{{36}} = \dfrac{1}{9}$$
$${\text{Let }}B{\text{ be the event that the sum of numbers rolle dover dice is 7,}}$$
$${\text{The favourable cases will be (1,6),(6,1),(2,5),(5,2),(4,3),(3,4) and total number of cases = 6}} \times {\text{6 = 36}}$$
$${\therefore\text{ } }P(B) = \dfrac{6}{{36}} = \dfrac{1}{6}$$
$${\text{Let }}E{\text{ be the event that the sum of the numbers rolled over dice is either 5 or 7,}}$$
$$P(E) = \dfrac{1}{9} + \dfrac{1}{6} = \dfrac{{2 + 3}}{{18}} = \dfrac{5}{{18}}$$
$${\text{So the probability of the event that the sum is neither 5 nor 7 would be,}}$$
$$P(\bar E) = 1 - \dfrac{5}{{18}} = \dfrac{{13}}{{18}}$$
$${\textbf{Step - 2: Find the probability that sum of 5 comes before 7}}{\text{.}}$$
$${\text{Let }}F{\text{ be the event that on rolling the dice sum of 5 comes before the sum of 7}}{\text{.}}$$
$$P(F) = P(A) + P(\bar E) \times P(A) + P(\bar E) \times P(\bar E) \times P(A) + .......\infty $$
$$P(F) = \dfrac{1}{9} + \dfrac{{13}}{{18}} \times \dfrac{1}{9} + \dfrac{{13}}{{18}} \times \dfrac{{13}}{{18}} \times \dfrac{1}{9} + ......\infty $$
$${\text{As we can see that it forms a }}G.P.{\text{we can find its sum,}}$$
$${\text{Sum of a }}G.P.{\text{ with first term }}a{\text{ and a common difference }}r,{\text{its sum upto }}$$
$${\text{infinite terms, can be written as}}$$
$$S = \dfrac{a}{{1 - r}}$$
$${\text{Here we have }}a = \dfrac{1}{9},r = \dfrac{{13}}{{18}}$$
$$S = \dfrac{{\dfrac{1}{9}}}{{1 - \dfrac{{13}}{{18}}}} = \dfrac{{\dfrac{1}{9}}}{{\dfrac{5}{{18}}}} = \dfrac{2}{5}.$$
$$\textbf{Hence option D is correct.}$$
Question 8
1 / -0

In throwing a pair of dice, the events 'coming up of 6 on Ist dice' and 'a total of 7 on both the dice' are

A
mutually exclusive

B
forming an exhaustive system

C
independent

D
dependent

Solution
Question 9
1 / -0

Two probability distributions of the discrete random variable $$X$$ and $$Y$$ are given below.

$$X$$ $$0$$ $$1$$ $$2$$ 3

$$P\left( X

\right)$$
$$\dfrac { 1 }{ 5

} $$ $$\dfrac { 2 }{ 5

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

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

$$Y$$ $$0$$ $$1$$ $$2$$ $$3$$

$$P\left( Y

\right)$$ $$\dfrac { 1 }{ 5 }$$ $$\dfrac { 3 }{ 10

} $$ $$\dfrac { 2 }{ 5

} $$ $$\dfrac { 1 }{ 10 }$$
Then

A
$$E\left( { Y }^{ 2 } \right) =2E\left( X \right)$$

B
$$E\left( { Y }^{ 2 } \right) =E\left( X \right)$$

C
$$E\left( Y \right) =E\left( X \right)$$

D
$$E\left( { X }^{ 2 } \right) =2E\left( Y \right) $$

Solution

$$E(X)=\sum { XP( } X)=0*\frac { 1 }{ 5 } +1*\frac { 2 }{ 5 } +2*\frac { 1 }{ 5 } +3*\frac { 1 }{ 5 } =7/5\\ E({ Y }^{ 2 })=\sum { { y }^{ 2 }P(Y) } ={ 0 }^{ 2 }*\frac { 1 }{ 5 } +1^{ 2 }*\frac { 3 }{ 10 } +{ 2 }^{ 2 }*\frac { 4 }{ 10 } +{ 3 }^{ 2 }*\frac { 1 }{ 10 } =28/10=14/5\\ E({ Y }^{ 2 })=2E(X)$$
Question 10
1 / -0

Let $$A$$ and $$B$$ be two finite sets having $$m$$ and $$n$$ elements respectively such that $$m \le n$$. A mapping is selected at random from the set of all mappings from $$A$$ to $$B$$. The probability that the mapping selected is an injection, is

A
$$\dfrac{n!}{(n-m)!m^{n}}$$

B
$$\dfrac{n!}{(n-m)!n^{m}}$$

C
$$\dfrac{m!}{(n-m)!n^{m}}$$

D
$$\dfrac{m!}{(n-m)!m^{n}}$$

Solution

A has m elements and B has n elements
Total number of mapping from A to B =$${ m }^{ n }$$
Injection mapping every element of A will be having distinct value in B so it is the total number of ways to choose m elements from n elements with permutation=$${ { n }_{ C } }_{ m }$$*m!
Probability=$${ { n }_{ C } }_{ m }$$*m!/$${ m }^{ n }$$=$$\frac { n! }{ (n-m)!{ n }^{ m } } $$

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Re-Attempt Weekly Quiz Competition

$$X$$	$$0$$	$$1$$	$$2$$	3
$$P\left( X \right)$$	$$\dfrac { 1 }{ 5 } $$	$$\dfrac { 2 }{ 5 } $$	$$\dfrac { 1 }{ 5 }$$	$$\dfrac { 1 }{ 5 }$$

$$Y$$	$$0$$	$$1$$	$$2$$	$$3$$
$$P\left( Y \right)$$	$$\dfrac { 1 }{ 5 }$$	$$\dfrac { 3 }{ 10 } $$	$$\dfrac { 2 }{ 5 } $$	$$\dfrac { 1 }{ 10 }$$

Probability Test - 54

Result

Probability Test - 54

Score

-

Rank

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SHARING IS CARING

Question 1

$$0.92$$

$$0.66$$

$$0.45$$

$$0.08$$

Solution

Question 2

$$2$$

$$log_2 e$$

$$log_e 2$$

4

Solution

Question 3

$$\dfrac{8}{27}$$

$$\dfrac{19}{27}$$

$$\dfrac{10}{27}$$

$$\dfrac{17}{27}$$

Solution

Question 4

$$0.657$$

$$0.965$$

$$0.796$$

$$0.509$$

Solution

Question 5

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

$$\frac{4}{9}$$

$$\frac{8}{27}$$

$$\frac{16}{81}$$

Solution

Question 6

3/12341

2/12341

1/12341

None of these

Question 7

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

$$\frac{1}{5}$$

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

$$\frac{2}{5}$$

Solution

Question 8

mutually exclusive

forming an exhaustive system

independent

dependent

Solution

Question 9

$$E\left( { Y }^{ 2 } \right) =2E\left( X \right)$$

$$E\left( { Y }^{ 2 } \right) =E\left( X \right)$$

$$E\left( Y \right) =E\left( X \right)$$

$$E\left( { X }^{ 2 } \right) =2E\left( Y \right) $$

Solution

Question 10

$$\dfrac{n!}{(n-m)!m^{n}}$$

$$\dfrac{n!}{(n-m)!n^{m}}$$

$$\dfrac{m!}{(n-m)!n^{m}}$$

$$\dfrac{m!}{(n-m)!m^{n}}$$

Solution

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