Probability Test

Result

Probability Test - 53

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

There are three boxes, each containing a different number of light bulbs. The first box has 10 bulbs, of which four are dead, the second box has six bulbs, of which one is dead, and the third box has eight bulbs of which three are dead. What is the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes?

A
$$\dfrac{115}{330}$$

B
$$\dfrac{113}{360}$$

C
$$\dfrac{113}{330}$$

D
None of these

Solution

Let $$𝐴_1, 𝐴_2, 𝐴_3$$ denotes the events of selecting bulbs from bags $$1,2\: and\: 3$$ respectively.

Let $$𝐵$$ denotes the event the bulbs selected are dead.

$$ 𝑃(𝐴_1) = 𝑃(𝐴_2) = 𝑃(𝐴_3) = \dfrac{1}{3} $$

Also $$P(B|A_1)=\dfrac{4}{10}, P(B|A_2)=\dfrac{1}{6}, P(B|A_3)=\dfrac{3}{8}$$

By law of total probability,

$$P(B)=P(A_1)P(B|A_1)+P(A_2)P(B|A_2)+P(A_3)P(B|A_3)$$

Substituting the values we get,

$$P(B)=\dfrac{1}{3} \times \dfrac{4}{10}+ \dfrac{1}{3} \times \dfrac{1}{6}+ \dfrac{1}{3} \times \dfrac{3}{8}$$

$$\Rightarrow P(B)=\dfrac{113}{360}$$

Thus the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes is $$\dfrac{113}{360}$$.
Question 2
1 / -0

Suppose that two factories supply light bulbs to the market. Factory X's bulbs work for over $$5000$$ hours in $$99\%$$ of cases, whereas factory Y's bulbs work for over $$5000$$ hours in $$95\%$$ of cases. It is known that factory X supplies $$60\%$$ of the total bulbs available. What is the chance that a purchased bulb will work for longer than $$5000$$ hours?

A
$$\dfrac{876}{1000}$$

B
$$\dfrac{544}{1000}$$

C
$$\dfrac{974}{1000}$$

D
None of these

Solution

Let $$X$$ be the event "comes from factory $$X$$" and $$Y$$ be the event "comes fom factory $$Y$$ " and Let $$H$$ be the event "works over $$5000$$ hours."

Therefore $$P(X)=60 \%=0.60 \Rightarrow P(Y)=1-0.60=0.40$$

Given that, Factory $$X's$$ bulbs work for over $$5000$$ hours in $$99 \%$$ of cases.

$$\therefore P(H|X)=99 \%=0.99$$

Also given, factory Y's bulbs work for over $$5000$$ hours in $$95\%$$ of cases.

$$\therefore P(H|Y)=95 \%=0.95$$

Then by the Law of Total Probability we have

$$P(H) = P (H | X) P(X) + P (H | Y ) P(Y ) $$

$$= (.99) (.6) + (.95) (.4)$$

$$ = .974$$

Thus $$P(H)=0.974=\dfrac{974}{1000}$$.
Question 3
1 / -0

If $$X$$ is a discrete random variable then which of the following is correct?

A
$$0\le F(x)<1$$

B
$$F(-\infty )=0;F(\infty )\le 1$$

C
$$P\left[ X={ x }_{ n } \right] =F({ x }_{ n })-F({ x }_{ n-1 })$$

D
$$F(x)$$ is a constant function

Solution
Question 4
1 / -0

A letter is known to have come either from TATANAGAR or from CALCUTTA. On the envelope just two Consecutive letters TA are visible. What is the probability that the letters came from TATANAGAR ?

A
$$\dfrac{7}{11}$$

B
$$\dfrac{24}{29}$$

C
$$\dfrac{4}{11}$$

D
$$\dfrac{1}{11}$$

Solution

Let $$E_1$$ denote the event that the letter came from TATANAGAR and $$E_2$$ denote the event that the letter came from CALCUTTA.

Let $$A$$ be the event that the two consecutive alphabets visible on the envelope are TA.

Since the letters have come from either Calcutta or Tatanagar.

Therefore, $$P(E_1)=\dfrac{1}{2}, P(E_2)=\dfrac{1}{2}$$

If $$E_1$$ has occurred then the letter has come from TATANAGAR. In the word TATANAGAR there are $$8$$ consecutive alphabets $$i.e.,\{TA,AT,TA,AN,NA,AG,GA,AR\}$$ and TA occurs two times.

$$\therefore P(A/E_1)=\dfrac{2}{8}=\dfrac{1}{4}$$

If $$E_2$$ has occurred then the letter has come from CALCUTTA. In the word CALCUTTA there are $$7$$ consecutive alphabets $$i.e.,\{CA,AL,LC,CU,UT,TT,TA\}$$ and TA occurs once.

$$\therefore P(A/E_2)=\dfrac{1}{7}$$

The probability that the letter has come from TATANAGAR is $$P(E_1/A)$$

By Bayes theorem, $$P(E_1/A)=\dfrac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}$$

$$=\dfrac{\dfrac{1}{2}\times \dfrac{1}{4}}{\dfrac{1}{2}\times\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{7}}$$

$$=\dfrac{\dfrac{1}{8}}{\dfrac{1}{8}+\dfrac{1}{14}}$$

$$\therefore P(E_1/A)=\dfrac{7}{11}$$

Hence the probability that the letter has come from TATANAGAR is $$\dfrac{7}{11}$$.
Question 5
1 / -0

For the events $$A$$ and $$B, P(A) = \dfrac {3}{4}, P(B) = \dfrac {1}{5}, P(A\cap B) = \dfrac {1}{20}$$ then $$P(A/B) =$$ ___________.

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

B
$$\dfrac {1}{15}$$

C
$$\dfrac {3}{4}$$

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

Solution

Given $$P(A)=\dfrac{3}{4}$$, $$P(B)=\dfrac{1}{5}$$, $$P(A\cap B)=\dfrac{1}{20}$$.

We have to find $$P(A/B)$$.

We know that $$P(A/B)=\dfrac{P(A\cap B)}{P(B)}$$.

Substituting the values we get

$$P(A/B)=\dfrac{\dfrac{1}{20}}{\dfrac{1}{5}}$$

$$=\dfrac{5}{20}$$

$$\therefore P(A/B)=\dfrac{1}{4}$$
Question 6
1 / -0

Directions For Questions

A bag contains 6 balls of 3 different colours namely White, Green and Red, atleast one ball of each different colour. Assume all possible probability distributions are equally likely.

...view full instructions

The probability that the bag contains 2 balls of each colour, is

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

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

C
$$\dfrac{1}{10}$$

D
$$\dfrac{1}{4}$$
Question 7
1 / -0

A random variable $$X$$ has the following probability distribution:

$$X$$ 0 1 2 3 4 5
$$P(X=x)$$ $$\cfrac{1}{4}$$ $$2a$$ $$3a$$ $$4a$$ $$5a$$ $$\cfrac{1}{4}$$
Then $$P(1\le X\le 4)$$ is:

A
$$\cfrac { 10 }{ 21 } $$

B
$$\cfrac { 2 }{ 7 } $$

C
$$\cfrac { 1 }{ 14 } $$

D
$$\cfrac { 1 }{ 2 } $$

Solution

We know that $$\sum P(X=x)=1$$

$$\Rightarrow \dfrac{1}{4}+2a+3a+4a+5a+\dfrac{1}{4}=1$$

$$\Rightarrow \dfrac{1}{2}+14a=1$$

$$\Rightarrow 14a=1-\dfrac{1}{2}$$

$$\Rightarrow 14a=\dfrac{1}{2}$$

$$\Rightarrow a=\dfrac{1}{28}$$

Now we have to find $$P(1\leq x \leq 4)$$

$$P(1\leq x \leq 4)=P(x=1)+P(x=2)+P(x=3)+P(x=4)$$

$$=2a+3a+4a+5a$$

$$=14a$$

$$=14 \left(\dfrac{1}{28}\right) \quad \left[\because a=\dfrac{1}{28}\right]$$

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

$$\therefore P(1\leq x \leq 4)=\dfrac{1}{2}$$
Question 8
1 / -0

The distribution of a random variable X is given below:
X = x 2 -1 0 1 2 3
P(X = x) $$\frac{1}{10}$$ k $$\frac{1}{5}$$ 2k $$\frac{3}{10}$$ k

A
$$\frac{1}{10}$$

B
$$\frac{2}{10}$$

C
$$\frac{3}{10}$$

D
$$\frac{7}{10}$$

Solution

Question 9

1 / -0

Let the p.m.f. of a random variable $$X$$ be -
$$P(x) = \dfrac {3 - x}{10}$$ for $$x = -1, 0, 1, 2 $$ otherwise
Then $$E(X)$$ is _________.

$$1$$

$$2$$

$$0$$

$$-1$$

Solution

$$x$$	$$-1$$	$$0$$	$$1$$	$$2$$
$$P(x)$$	$$\dfrac{4}{10}$$	$$\dfrac{3}{10}$$	$$\dfrac{2}{10}$$	$$\dfrac{1}{10}$$
$$x\cdot P(x)$$	$$-\dfrac{4}{10}$$	$$0$$	$$\dfrac{2}{10}$$	$$\dfrac{2}{10}$$

We have to find $$E(X)$$

We know that $$E(X)=\sum x\cdot P(x)$$

$$\therefore E(X)=-\dfrac{4}{10}+0+\dfrac{2}{10}+\dfrac{2}{10}=0$$

Hence $$E(X)=0$$

Question 10
1 / -0

There is a probability that 4 out of 10 students appearing in CA CPT exams will qualify for CA course after passing CA CPT exam, the probability that of 5 students none will join CA course is.......

A
$$0.01$$

B
$$0.065$$

C
$$0.43$$

D
$$40.08$$

Solution

Total no. students $$=10$$
$$\Rightarrow$$ Probability of students will qualify $$=\dfrac{4}{10}$$
$$\Rightarrow$$ Probability of students who will not qualify $$=\dfrac{6}{10}$$
i.e, $$p=\dfrac{4}{10}$$ $$q=\dfrac{6}{10}$$ $$r=5$$
Probability that out of $$5$$ students none will join CA
By Bemoulli's equation,
$$\Rightarrow {^nC_rp^rq^{n-r}}={^5C_0}\left( \dfrac { 4 }{ 10 } \right) ^{ 0 }\left( \dfrac {6 }{ 10 } \right) ^{ 5 }=0.077\sim 0.08$$
Hence, the answer is $$0.08.$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$X$$	0	1	2	3	4	5
$$P(X=x)$$	$$\cfrac{1}{4}$$	$$2a$$	$$3a$$	$$4a$$	$$5a$$	$$\cfrac{1}{4}$$

X = x	2	-1	0	1	2	3
P(X = x)	$$\frac{1}{10}$$	k	$$\frac{1}{5}$$	2k	$$\frac{3}{10}$$	k

Probability Test - 53

Result

Probability Test - 53

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{115}{330}$$

$$\dfrac{113}{360}$$

$$\dfrac{113}{330}$$

None of these

Solution

Question 2

$$\dfrac{876}{1000}$$

$$\dfrac{544}{1000}$$

$$\dfrac{974}{1000}$$

None of these

Solution

Question 3

$$0\le F(x)<1$$

$$F(-\infty )=0;F(\infty )\le 1$$

$$P\left[ X={ x }_{ n } \right] =F({ x }_{ n })-F({ x }_{ n-1 })$$

$$F(x)$$ is a constant function

Solution

Question 4

$$\dfrac{7}{11}$$

$$\dfrac{24}{29}$$

$$\dfrac{4}{11}$$

$$\dfrac{1}{11}$$

Solution

Question 5

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

$$\dfrac {1}{15}$$

$$\dfrac {3}{4}$$

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

Solution

Question 6

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

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

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

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

Question 7

$$\cfrac { 10 }{ 21 } $$

$$\cfrac { 2 }{ 7 } $$

$$\cfrac { 1 }{ 14 } $$

$$\cfrac { 1 }{ 2 } $$

Solution

Question 8

$$\frac{1}{10}$$

$$\frac{2}{10}$$

$$\frac{3}{10}$$

$$\frac{7}{10}$$

Solution

Question 9

$$1$$

$$2$$

$$0$$

$$-1$$

Solution

Question 10

$$0.01$$

$$0.065$$

$$0.43$$

$$40.08$$

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.