Probability Test

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Probability Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

$$10$$ mangoes are to be distributed among $$5$$ persons. The probability that at least one of them will receive none is :

A
$$\dfrac{35}{143}$$

B
$$\dfrac{108}{143}$$

C
$$\dfrac{18}{143}$$

D
$$\dfrac{125}{143}$$

Solution

10 mangoes can be distributed among 5 persons in
$$^{10+5- 1}C_{5-1}= ^{14} C_4$$ ways
Therefore, Total number of elementary events =$$^{14}C_4$$
Required probability
= 1 – probability that each person received at least
one mango
$$\implies = 1-\dfrac {^{10-1}C_{5-1}}{^{14}C_4}\\\implies 1-\dfrac {^9C_4}{^{14}C_4}\\\implies =1-\dfrac {9!}{4!5!}\times \dfrac {4!10!}{14!}=1-\dfrac {18}{143}\\\implies = \dfrac {125}{143}$$
Question 2
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Which of the following is true regarding law of total probability?

A
It is a fundamental rule relating marginal probabilities to conditional probabilities.

B
It expresses the total probability of an outcome which can be realized via several distinct events

C
Both are correct

D
None of these

Solution

Law of total probability states that,
If $$B_1,B_2,B_3,...$$ is a partition of the sample space $$S$$, then for any event $$A$$ we have
$$P(A)=\sum _iP(A\cap B_i)=\sum_i P(A|B_i)P(B_i)$$

That is, the law (or formula) of total probability is a fundamental rule relating marginal probabilities to conditional probabilities. It expresses the total probability of an outcome which can be realized via several distinct events and hence the name.

Thus option $$(C)$$ is correct.
Question 3
1 / -0

From $$30$$ bulbs out of which $$10$$ are defective, $$3$$ bulbs are chosen. X denotes number of defective bulbs.The PDF of r.v x is
x 0 1 2 3
p(x) 57k 95k 45k 6k
then k is

A
$$\dfrac{1}{210}$$

B
$$\dfrac{2}{203}$$

C
$$\dfrac{1}{203}$$

D
$$\dfrac{1}{29}$$
Solution
We know that, Suppose a random variable $$X$$ may take $$k$$ different values, with the probability that $$X=x_i$$ defined to be $$P(X=x_i)=p_i$$. The probabilities $$p_i$$ must satify the following:
1. $$0\leq p_i\leq1$$ for each $$i$$.
2. $$p_1+p_2+...+p_k=1$$ that is $$\sum p_i=1$$
Thus we have $$\sum p_i=1$$

$$\Rightarrow p_0+p_1+p_2+p_3=1$$

$$\Rightarrow 57k+95k+45k+6k=1$$

$$\Rightarrow 203k=1$$

$$\Rightarrow k=\dfrac{1}{203}$$
Question 4
1 / -0

A bag contains three red balls and two blue balls. We remove successively, and with replacement, two balls and observe their colour. What is the probability of "to extract a red ball and a blue one, without taking into account the order"?

A
$$\dfrac{12}{25}$$

B
$$\dfrac{13}{25}$$

C
$$\dfrac{12}{23}$$

D
None of these

Solution

Given:
Total number of balls in a bag = 3 red balls + 2 blue balls = 5
Let's consider the events: R= "to remove a red ball", A=R'= "to remove a blue ball.
Our sample space is: (RR, RA, AR, AA}, and the events that we are interested in are RA and AR.
We represent our problem in a tree as shown in the figure
Using the law of total probability,
$$P(S)=P(R)\times P(A/R)+P(A)\times P(R/A)=\dfrac 35\times \dfrac 25+\dfrac 25\times \dfrac 35=\dfrac {12}{25}$$
is the required probability.
Question 5
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One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. What is the probability that both the balls will be of the same colour?

A
$$207/625$$

B
$$191/625$$

C
$$23/625$$

D
$$227/625$$

Solution

$$\textbf{Step-1: Find the number of balls in each bag.}$$

$$\text{Bag}$$ $$I$$ $$\text{ contains 3 white, 7 red and 15 black balls.}$$

$$\text{Bag}$$ $$II$$ $$\text{contains 10white, 6 red and 9 black balls.}$$

$$\therefore$$ $$\text{Each bag contains total of}$$ $$ 25$$ $$\text{balls.}$$

$$\textbf{Step-2: Find the number of cases for the selection of a ball.}$$

$$\text{There are three cases for selection of a particular ball:-}$$

$$\text{Case 1:}$$ $$\text{Probability of getting one white ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{3}{25}\times\dfrac{10}{25}$$

$$\text{Case 2:}$$ $$\text{Probability of getting one red ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{7}{25}\times\dfrac{6}{25}$$

$$\text{Case 3:}$$ $$\text{Probability of getting one black ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{15}{25}\times\dfrac{9}{25}$$

$$\textbf{Step-3: Add the above cases to get the required probability.}$$

$$\therefore$$ $$\text{Total probability that both the ball will be of same colour}$$ $$=\dfrac{30} {625}+\dfrac{42}{625}+\dfrac{135}{25}=\dfrac{207}{625}.$$

$$\textbf{Hence, the answer is}$$ $$\boldsymbol{\dfrac{207}{625}}.$$
Question 6
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I have three bags that each contain $$100$$ marbles- Bag $$1$$ has $$75$$ red and $$25$$ blue marbles, Bag $$2$$ has $$60$$ red and $$40$$ blue marbles, Bag $$3$$ has $$45$$ red and $$55$$ blue marbles. I choose one of the bags at random and then pick a marble from the chosen bag, also at random. What is the probability that the chosen marble is red?

A
$$0.60$$

B
$$0.40$$

C
$$0.50$$

D
None of these

Solution

Given that bag 1 contains $$75$$ red and $$25$$ blue marbles
Given that bag 2 contains $$60$$ red and $$40$$ blue marbles
Given that bag 3 contains $$45$$ red abd $$55$$ blue marbles
Now the probability of choosing red ball from bag 1 is $$ \dfrac{1}{3} \times \dfrac{75}{100}$$
Now the probability of choosing red ball from bag 2 is $$ \dfrac{1}{3} \times \dfrac{60}{100}$$
Now the probability of choosing red ball from bag 3 is $$ \dfrac{1}{3} \times \dfrac{45}{100}$$
Now the probability of choosing red ball is $$ \dfrac{1}{3}\left(\dfrac{75+60+45}{100}\right)=0.6$$
Hence, option $$A$$ is correct.
Question 7
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We draw two cards from a deck of shuffled cards without replacement. Find the probability of getting the second card a queen.

A
$$\dfrac{1}{13}$$

B
$$\dfrac{2}{13}$$

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

D
None of these

Solution

There are two cases here:
Case 1: First card chosen is a queen
$$\dfrac 4{52}\times \dfrac 3{51}=\dfrac {1}{221}$$
Case 2: First card chosen is not a queen.
$$\dfrac {48}{52}\times \dfrac 4{51}=\dfrac {16}{221}$$
Adding both the cases, we get the probability of getting the second card a queen. $$\dfrac {17}{221} =\dfrac 4{52} = \dfrac 1{13}$$
Question 8
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At a certain university, $$4$$% of men are over $$6$$ feet tall and $$1$$% of women are over 6 feet tall. The total student population is divided in the ratio $$3:2$$ in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?

A
$$\dfrac{3}{11}$$

B
$$\dfrac{5}{11}$$

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

D
none of these

Solution

Let M={Student is Male}, F={Student is Female}.
Note that M and F partition the sample space of students.
Let T={Student is over 6 feet tall}.
We know that $$P(M) =\dfrac 25, P(F) =\dfrac 35, P(T/M) = \dfrac 4{100}$$ and $$P(T/F) = \dfrac 1{100}$$.
We require $$P(F/T)$$.
Using Bayes’ theorem we have:
$$P(F/T) = \dfrac {P(T/F)P(F)}{P(T/F)P(F) + P(T/M)P(M)}\\\implies =\dfrac {\dfrac 1{100}\times \dfrac 35}{\dfrac 1{100}\times \dfrac 35+\dfrac 4{100}\times \dfrac 25}=\dfrac {3}{500}\times \dfrac {500}{3+8}=\dfrac {3}{11}$$
Question 9
1 / -0

Two coins are tossed. Find the conditional probability that two Heads will occur given that at least one occurs.

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

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

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

D
none of these

Solution

Let A = {at least one Head occurs}, B = {two Heads occur}
Given:
Two coins are tossed.
To find:
The conditional probability that two Heads will occur given that at least one occurs.
$$P(B) = \dfrac {P(A\cap B)}{P(A\cup B)}=\dfrac {P(A) \times P(B)}{P(A)+P(B)-P(A)\times P(B)}=\dfrac {\dfrac 12\times \dfrac 12}{\dfrac 12+\dfrac 12-\dfrac 12\times \dfrac 12}=\dfrac 14\times \dfrac {4}{2+2-1}=\dfrac 13$$
Question 10
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From a batch of $$100$$ items of which $$20$$ are defective, exactly two items are chosen, one at a time, without replacement. Calculate the probabilities that the second item chosen is defective.

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

B
$$\dfrac{19}{100}$$

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

D
none of these

Solution

Let A={first item chosen is defective}, B ={second item chosen is defective}
Given:
Total number of items = 100.
Number of defective items = 20
To find:
the probabilities that second item chosen is defective.
$$P(B)=P(B/A)P(A)+P(B/A')P(A')=\dfrac {20}{100}\times \dfrac {19}{99}+\dfrac {20}{99}\times \dfrac {80}{100}=\dfrac {198}{990}=\dfrac 15$$

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Probability Test - 28

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Probability Test - 28

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SHARING IS CARING

Question 1

$$\dfrac{35}{143}$$

$$\dfrac{108}{143}$$

$$\dfrac{18}{143}$$

$$\dfrac{125}{143}$$

Solution

Question 2

It is a fundamental rule relating marginal probabilities to conditional probabilities.

It expresses the total probability of an outcome which can be realized via several distinct events

Both are correct

None of these

Solution

Question 3

$$\dfrac{1}{210}$$

$$\dfrac{2}{203}$$

$$\dfrac{1}{203}$$

$$\dfrac{1}{29}$$

Solution

Question 4

$$\dfrac{12}{25}$$

$$\dfrac{13}{25}$$

$$\dfrac{12}{23}$$

None of these

Solution

Question 5

$$207/625$$

$$191/625$$

$$23/625$$

$$227/625$$

Solution

Question 6

$$0.60$$

$$0.40$$

$$0.50$$

None of these

Solution

Question 7

$$\dfrac{1}{13}$$

$$\dfrac{2}{13}$$

$$\dfrac{5}{13}$$

None of these

Solution

Question 8

$$\dfrac{3}{11}$$

$$\dfrac{5}{11}$$

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

none of these

Solution

Question 9

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

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

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

none of these

Solution

Question 10

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

$$\dfrac{19}{100}$$

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

none of these

Solution

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