Probability Test

Result

Probability Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

A class consists of $$n$$ students. For $$0\leq k\leq n$$, let $$E_k$$ denote the event that exactly $$k$$ student out of $$n$$ pass in the examination. Let $$P(E_k)=p_k$$ and let $$A$$ denote the event that a student $$X$$ selected at random pass in the examination.

...view full instructions

If $$P(E_k)=C$$ for $$0\leq k\leq n$$, then $$P(A)$$ equals

A
$$1/2$$

B
$$2/3$$

C
$$1/6$$

D
$$1/(n+1)$$

Solution

$$P(E_k)=C$$ for $$0\leq k\leq n$$ and $$\displaystyle \sum_{k=0}^nP(E_k)=1\Rightarrow C=1/(n+1)$$.
Also $$\displaystyle P(A|E_k)=\frac {^{n-1}C_{k-1}}{^nC_k}=\frac {k}{n}$$
By the total probability rule
$$P(A)=\displaystyle \sum_{k=0}^nP(E_k)P(A|E_k)$$
$$\displaystyle =\frac {1}{n(n+1)}\frac {n(n+1)}{2}=\frac {1}{2}$$.
Question 2
1 / -0

A box contain $$N$$ coins, $$m$$ of which are fair are rest and biased. The probability of getting a head when a fair coin is tossed is $$1/2$$, while it is $$2/3$$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

A
$$\dfrac {9m}{8N+m}$$

B
$$\dfrac {m}{8N+m}$$

C
$$\dfrac {N}{8n+m}$$

D
$$\dfrac {1}{m}$$

Solution

Let $$E_1, E_2$$ and $$A$$ denote the following events:
$$E_1$$: coin selected is fair
$$E_2$$: coin selected is biased
$$A: $$ the first toss results in a head and the second toss results in a tail.
$$\displaystyle P(E_1)=\frac {m}{N}, P(E_2)=\frac {N-m}{N}$$,
$$\displaystyle P(A|E_1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E_2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$$.
By Bayes' rule
$$\displaystyle P(E_1|A)=\frac {P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\frac {9m}{8N+m}$$.
Question 3
1 / -0

A player throws a fair cubical die and score the number appearing on the die. If he throws a $$1$$, he gets a further throw. Let $$p_r$$ denote the probability of getting a total score of exactly $$r$$. Sum of the series $$\displaystyle S=\sum_{r=1}^{\infty}p_r$$ is

A
1

B
1/6

C
1/5

D
2/3

Solution

Let $$P_r$$ denote the probability of getting a total score of exactly.
$$\Rightarrow A=$$ Number appearing on the die.
$$\Rightarrow B=$$ He throws a one and get a further throw
      $$=P\left( \dfrac { A }{B } \right)$$
          $$=$$ Probability of occurrence of $$A$$ when $$B$$ is taken as the sample space,
          $$=\dfrac{No.\;of\;elementary\;events\;favorable\;to\;A\;which\;are\;favorable\;to\;B}{No.\;of\;elementory\;events\;favorable\;to\;B}$$
          $$=\dfrac{No.\;of\;elements\;events\;favorable\;to\;(A\cap B)}{No.\;of\;elementory\;events\;favorable\;to\;B}$$
          $$=\dfrac{1}{5}.$$
Hence, the answer is $$\dfrac{1}{5}.$$
Question 4
1 / -0

A bag contains $$(2n+1)$$ coins. It is known that $$n$$ of these coins have a head on both sides, whereas the remaining $$n+1$$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $$\displaystyle \frac{31}{42}$$, then $$n$$ is equal to

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

Let $${ A }_{ 1 }$$ denote the event that a coin having heads on both sides is chosen, and $${ A }_{ 2 }$$ denote the vent that a fiar coin is chosen.
Let $$E$$ denote the vent that head occurs, Then
$$\displaystyle P\left( { A }_{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A }_{ 2 } \right) =\frac { n+1 }{ 2n+1 } $$
Probability of occurrence of event $$E$$, if unfair coin was selected is $$\displaystyle P\left( \frac { E }{ { A }_{ 1 } } \right) =1$$
Probability of occurrence of event $$E$$, if fair coin was selected is $$\displaystyle P\left( \frac { E }{ { A }_{ 2 } } \right) =\frac { 1 }{ 2 } $$
$$\because P\left( E \right) =P\left( { A }_{ 1 }\cap E \right) +P\left( { A }_{ 2 }\cap E \right) $$
$$\displaystyle \therefore P\left( E \right) =P\left( { A }_{ 1 } \right) P\left( \frac { E }{ { A }_{ 1 } } \right) +P\left( { A }_{ 2 } \right) P\left( \frac { E }{ { A }_{ 2 } } \right) $$
$$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right) } \\ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$$
Question 5
1 / -0

The probability that certain electronic component fails when first used is $$0.10.$$ If it does not fail immediately, the probability that is lasts for one year is $$0.99.$$ The probability that a new component will last for one year is

A
$$0.891$$

B
$$0.692$$

C
$$0.92$$

D
None of these

Solution

Given: probability that electronic component fails when first used $$\displaystyle =0.10,$$ i.e., $$\displaystyle P\left( F \right) =0.10$$
$$\displaystyle \therefore P\left( F' \right) =1-P\left( F \right) =0.90$$
and let $$P(Y)=$$ probability of new component to last for one year
Obviously, the two events are mutually exclusive and exhaustive.
$$\displaystyle \therefore P\left( \frac { Y }{ F } \right) =0$$ and $$\displaystyle P\left( \frac { Y }{ F' } \right) =0.99$$
$$\displaystyle \therefore P\left( Y \right) =P\left( F \right) .P\left( \frac { Y }{ F } \right) +P\left( F' \right) .P\left( \frac { Y }{ F' } \right) $$
$$\displaystyle =0.10\times 0+0.90\times 0.99=0+\left( 0.9 \right) \left( 0.99 \right) =0.891.$$
Question 6
1 / -0

If two events $$A$$ and $$B$$ are such that $$P\left( A' \right) =0.3,P\left( B \right) =0.4$$ and $$P\left( A\cap B' \right) =0.5$$, then $$\displaystyle P\left( \frac { B }{ A\cup B' } \right) $$ equals

A
$$\displaystyle \frac { 3 }{ 4 } $$

B
$$\displaystyle \frac { 5 }{ 6 } $$

C
$$\displaystyle \frac { 1 }{ 4 } $$

D
$$\displaystyle \frac { 3 }{ 7 } $$

Solution

$$\displaystyle P\left( \frac { B }{ A\cup B' } \right) =\frac { P\left[ B\cap \left( A\cap B' \right) \right] }{ P\left( A\cup B' \right) } =\frac { P\left[ \left( B\cap A \right) \cup \left( B\cap B' \right) \right] }{ P\left( A \right) +P\left( B' \right) -P\left( A\cap B' \right) } $$
But $$\displaystyle P\left[ \left( B\cap A \right) \cup \left( B\cap B' \right) \right] =P\left( B\cap A \right) =P\left( A \right) -P\left( A\cap B' \right) =0.2$$
and, $$\displaystyle P\left( A \right) +P\left( B' \right) -P\left( A\cap B' \right) =0.7+0.6-0.5=0.8$$
$$\displaystyle \therefore P\left( \frac { B }{ A\cup B' } \right) =\frac { 0.2 }{ 0.8 } =\frac { 1 }{ 4 } $$
Question 7
1 / -0

Let $$p$$ be the probability that a man aged $$x$$ years will die in a year time. The probability that out of $$n$$ men $$\displaystyle A_{1},A_{2},A_{3},....,A_{n}$$ each aged $$x$$ years, $$\displaystyle A_{1}$$ will die & will be the first to die, is

A
$$\displaystyle \frac{1-p^{n}}{n}$$

B
$$\displaystyle \frac{p}{n}$$

C
$$\displaystyle \frac{p\left ( 1-p \right )^{n-1}}{n}$$

D
$$\displaystyle \frac{1-\left ( 1-p \right )^{n}}{n}$$

Solution

Let $$E_i$$ be the event that $$A_i$$ will die in a year where i=1, 2, 3, 4, ..., n
The probability that none of $$A_1, A_2, A_3, ..., A_n $$ dies in a year
=$$(1-p_{A_1}).(1-p_{A_2}), ..., (1-p_{A_n})$$
Where $$p_{A_1}, p_{A_2}, ..., p_{A_n}$$ are the probabilities that $$A_1, A_2, A_3, ..., A_n $$ dies in a year respectively.
=$$(1-p).(1-p), ...(1-p)=(1-p)^n$$
Now the probability that at least one of $$A_1, A_2, A_3, ..., A_n $$
=$$1-(1-p)^n$$
The probability that among n men $$A_1$$ is the first one to die is $$\dfrac 1n$$
Therefore, the required probability is
$$=\dfrac 1n[1-(1-p)^n]$$
Question 8
1 / -0

A random variable $$X$$ has the probability distribution
$$X$$ 1 2 3 4 5 6 7 8
$$P(X)$$ 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
For the events, $$E = \{X$$ is a prime number$$\}$$ and $$F = {X < 4}$$, the probability $$\displaystyle P\left ( E\cup F \right )$$ is

A
$$0.87$$

B
$$0.77$$

C
$$0.35$$

D
$$0.50$$

Solution

From the given values, prime numbers and the values less than $$ 4 $$ are $$ 1,2,3,5,7 $$

So, $$ P(E \cup F) $$ will be the sum of probabilities of each of the above selected values of $$ X $$

So, $$ P( E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77 $$
Question 9
1 / -0

A fair coin is tossed $$99$$ times. If X is the number of times heads occur then P(X = r) is maximum when r is

A
$$49$$

B
$$50$$

C
$$51$$

D
none of these

Solution

Notice that x has binomial distribution with n = 99, p = q = 0.5.
Therefore the probability
$$P(x=r)=^nC_rP^rq^{n-r}\\\implies P(x=r)=^nC_r(0.5)^r(0.5)^{n-r}=^{n}C_r(0.5)^n$$
Thus maximum of P(x = r) corresponds to the value of r with maximal
$$^nC_r=\dfrac {n!}{r!(n-r)!}$$
Notice that binomial coefficients are symmetric:
$$^nC_r=^{n-r}C_r=\dfrac {n!}{r!(n-r)!}$$
hence the maximum corresponds to the points most closest to $$\dfrac n2$$
Since n is odd, there are two such values:
$$r_1=\dfrac {n-1}2=\dfrac {98}2=49, r_2=\dfrac {n+1}2=\dfrac {100}2=50$$
For these numbers the function P(x = r) has maximum.
Question 10
1 / -0

A purse contains $$2$$ six-sided dice. One is a normal fair die,while the other has two $$1's, $$ two $$ 3' s$$ and two $$5'$$, A die is picked up and rolled. Because of some secret magnetic attration of the unfair die, there is 75% chance of picking the unfair die and a 25% chance of picking a fair die. The die is rolled and shows up the face $$3$$ The probability that a fair die was picked up is

A
$$1/7$$

B
$$1/4$$

C
$$1/6$$

D
$$1/24$$

Solution

Let N be the normal die picked, and M be the magnetic die picked and also let A be die shows up 3.
According to the question,
$$P(N)=\dfrac 14, P(M)=\dfrac 34$$
$$P(A)=P(A\cap N)+P(A\cap M)\\\implies P(A)=P(N)\times P\left( \dfrac AN\right)+P(M)\times P\left(\dfrac AM\right)\\\implies P(A)=\dfrac 14\times \dfrac 16+\dfrac 34\times \dfrac 13=\dfrac 1{24}+\dfrac 14=\dfrac 7{24}$$

$$P\left(\dfrac NA\right)=\dfrac {P(N\cap A)}{P(A)}=\dfrac {\dfrac 14\times \dfrac 16}{\dfrac 7{24}}=\dfrac 17$$
is the required probability.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$X$$	1	2	3	4	5	6	7	8
$$P(X)$$	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

Probability Test - 50

Result

Probability Test - 50

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1/2$$

$$2/3$$

$$1/6$$

$$1/(n+1)$$

Solution

Question 2

$$\dfrac {9m}{8N+m}$$

$$\dfrac {m}{8N+m}$$

$$\dfrac {N}{8n+m}$$

$$\dfrac {1}{m}$$

Solution

Question 3

1

1/6

1/5

2/3

Solution

Question 4

$$10$$

$$11$$

$$12$$

$$13$$

Solution

Question 5

$$0.891$$

$$0.692$$

$$0.92$$

None of these

Solution

Question 6

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

$$\displaystyle \frac { 5 }{ 6 } $$

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

$$\displaystyle \frac { 3 }{ 7 } $$

Solution

Question 7

$$\displaystyle \frac{1-p^{n}}{n}$$

$$\displaystyle \frac{p}{n}$$

$$\displaystyle \frac{p\left ( 1-p \right )^{n-1}}{n}$$

$$\displaystyle \frac{1-\left ( 1-p \right )^{n}}{n}$$

Solution

Question 8

$$0.87$$

$$0.77$$

$$0.35$$

$$0.50$$

Solution

Question 9

$$49$$

$$50$$

$$51$$

none of these

Solution

Question 10

$$1/7$$

$$1/4$$

$$1/6$$

$$1/24$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.