Probability Test

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Probability Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

An experiment can result in only 3 mutually exclusive events A, B and C. If $$ \displaystyle P\left ( A \right )=2P\left ( B \right )=3P\left ( C \right ), $$ then $$ \displaystyle P\left ( A \right ) $$ equals

A
$$ \displaystyle \frac{5}{11} $$

B
$$ \displaystyle \frac{9}{11} $$

C
$$ \displaystyle \frac{8}{11} $$

D
$$ \displaystyle \frac{6}{11} $$

Solution

since the events A,B and C are mutually exclusive
Hence $$P\left( A\cap B \right) =P\left( B\cap C \right) =P\left( C\cap A \right) =P\left( A\cap B\cap C \right) =0\\ Also\quad P\left( A\cup B\cup C \right) =1$$
$$\because P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) -P\left( C\cap A \right) -P\left( A\cap B\cap C \right) \\ \Rightarrow 1=P\left( A \right) +\dfrac { P\left( A \right) }{ 2 } +\dfrac { P\left( A \right) }{ 3 } \\ \Rightarrow \dfrac { 11 }{ 6 } P\left( A \right) =1\\ \Rightarrow P\left( A \right) =\dfrac { 6 }{ 11 } $$
Question 2
1 / -0

A die is thrown :
$$P$$ is the event of getting an odd number.
$$Q$$ is the event of getting an even number.
$$R$$ is the event of getting a prime number.
Which of the following pairs is mutually exclusive?

A
$$P,Q$$

B
$$Q,R$$

C
$$P,R$$

D
None of these

Solution

Let $$S$$ be the sample space
$$S=\{1,2,3,4,5,6\}$$
$$n(S)=6$$
$$P$$ is the event of getting an odd number
$$\therefore P=\{1,3,5\}\implies n(P)=3$$
$$Q$$ is the event of getting an even number
$$\therefore Q=\{2,4,6\}\implies n(Q)=3$$
$$R$$ is the vent of getting a prime number
$$\therefore R=\{2,3,5\}\implies n(R)=3$$
The set $$S$$ contains two types of numbers even and odd
We have, $$P=S-Q$$ and $$Q=S-P$$
$$\therefore n(P)$$ and $$n(Q)$$ both are complementary.
Two events are mutually exclusive if they are disjoint
Now, $$P\cap Q=\phi$$
$$\implies P$$ and $$Q$$ are disjoint i.e. they don't have common elements.
Hence, $$P$$ and $$Q$$ are mutually exclusive.
Question 3
1 / -0

$$A$$ and $$ B$$ are events such that $$\displaystyle p(A \cup B) =3/4,P(A\cap B)=\frac{1}{4},P(\bar{A})=\frac{2}{3}$$, then $$\displaystyle P((\bar{A}\cap B)$$ equals

A
$$\displaystyle \frac{5}{12}$$

B
$$\displaystyle \frac{3}{8}$$

C
$$\displaystyle \frac{4}{5}$$

D
$$\displaystyle \frac{5}{4}$$

Solution

Given $$\displaystyle P(A\cup B)=3/4$$
$$\displaystyle P(A\cap B)=1/4$$ $$\displaystyle P(\bar{A})=2/3\therefore P(B)=\frac{2}{3}$$
By using $$\displaystyle P(B) =P(A\cup B)+P(A\cap B)-P(A)$$
$$\displaystyle\therefore P(\bar{A}\cap B)=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$$
Question 4
1 / -0

Form two digit numbers using the digit $$0,1,2,3,4,5$$ without repeating the digits.
$$P$$ is the event that the number so formed is even.
$$Q$$ is the event that the number so formed is divisible by $$3$$.
$$R$$ is the event that the number so formed is greater than $$50$$.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 5$$

B
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 2$$

C
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 4$$

D
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 8$$

Solution

The given digits are $$0,1,2,3,4,5$$
Number of two digit numbers that can be formed, as first place can be filed with 5 numbers and second place can be filed with another 5 numbers; $$5 \times 5 = 25$$
Now, number is an even number, with 0 at the end there can be $$5$$ numbers, with $$2$$ at the end there can be $$4$$ numbers and with $$4$$ at the end there can be $$4$$ numbers. Hence, total numbers which are even $$= 4 + 4 + 5 = 13$$
Numbers divisible by $$3$$ are, $$12, 15, 21, 24, 30, 42, 45, 51, 54$$
Thus, total numbers are, $$9$$
There are only $$4$$ number, viz. $$51, 52,53,54$$ which are greater than $$50$$
Hence, $$n(S) = 25, n(P) = 13, n(Q) = 9, n (R) = 4$$
Question 5
1 / -0

If $$P(A)\, =\, \displaystyle \frac{3}{4} ,\, P(B')\, =\, \frac{1}{3}\, and\, P(A \cap B)\, =\, \frac{1}{2}\,$$ then find $$P(A \cup B)$$.

A
$$\displaystyle \frac{5}{12}$$

B
$$\displaystyle \frac{6}{12}$$

C
$$\displaystyle \frac{7}{12}$$

D
$$\displaystyle \frac{11}{12}$$

Solution

Given that:
$$P(A)=\cfrac34, P(B')=\cfrac13$$ and $$P(A\cap B)=\cfrac12$$

To Find:
$$P(A\cup B)=?$$

Solution:
$$P(B)=1-P(B')=1-\cfrac13=\cfrac23$$

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

$$=\cfrac34+\cfrac23-\cfrac12$$

$$=\cfrac{11}{12}$$

Hence, D is the correct option.
Question 6
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There are $$3$$ men and $$2$$ women. a 'Gramswachhatta Abhiyan' committee of two is to be formed :
$$P$$ is event that the committee should contain at least one woman.
$$Q$$ is event that the committee should contain one man and one women.
$$R$$ is the event there should not be a women in the committee.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 10, n(P) = 7, n(Q) = 1, n(R) = 3$$

B
$$n(S) = 10, n(P) = 7, n(Q) = 3, n(R) = 3$$

C
$$n(S) = 10, n(P) = 7, n(Q) = 2, n(R) = 3$$

D
$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$

Solution

$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$
$$P$$ and $$ R$$ are complementary, mutually exclusive events and exhaustive events.
$$Q$$ and $$R$$ are complementary and mutually exclusive events.
Question 7
1 / -0

A card is drawn at random from well shuffled pack of $$52$$ cards. Find the probability that the card drawn is a spade:

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

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{2}$$

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

Solution

Sample space $$= 52$$
Event a card obtained is of spade $$= 13$$(spades ,$$13$$ cards of each suits)
Probability $$=\cfrac{13}{52}$$ $$=\cfrac{1}{4}$$
Question 8
1 / -0

A and B are two events on a sample space S such that $$P(A) = 0.8 , P(B) = 0.6$$ and $$P(A \cup B) = 0.9$$ find $$(A \cap B)$$.

A
0.3

B
0.4

C
0.5

D
0.6

Solution

Given that:
$$P(A)=0.8, P(B)=0.6$$ and $$P(A\cup B)=0.9$$
To find: $$P(A\cap B)=?$$
$$P(A\cap B)=P(A)+P(B)-P(A\cup B)$$
$$=0.8+0.6-0.9$$
$$=1.4-0.9$$
$$=0.5$$
Hence, C is the correct option.
Question 9
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There are $$3$$ red, $$3$$ white and $$3$$ green balls in a bag. One ball is drawn at random from a bag:
$$P$$ is the event that ball is red.
$$Q$$ is the event that ball is not green.
$$R$$ is the event that ball is red or white.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 9, n(P) =3, n(Q) = 6, n(R) = 6$$

B
$$n(S) = 4, n(P) =3, n(Q) = 6, n(R) = 6$$

C
$$n(S) = 6, n(P) =3, n(Q) = 6, n(R) = 6$$

D
$$n(S) = 7, n(P) =3, n(Q) = 6, n(R) = 6$$

Solution

There are $$3$$ red balls, $$3$$ white balls and $$3$$ green balls.
Thus, total number of balls $$= 9$$, $$n (S) = 9$$
$$P$$, the ball is red, $$n(R) = 3$$
$$Q$$, the ball is not green, $$n(Q) = 9 - 3 = 6$$
$$R$$, the ball is red or white, $$n(R) = 3 + 3 = 6$$
Question 10
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If two coins are tossed then find the probability of the event of getting one head .

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

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{2}$$

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

Solution

sample space is ={h,t} ,{t,h}
favourable outcomes for getting one head =$$2 $$
probability (one head)= $$\cfrac{2}{4}$$

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Probability Test - 29

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Probability Test - 29

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Question 1

$$ \displaystyle \frac{5}{11} $$

$$ \displaystyle \frac{9}{11} $$

$$ \displaystyle \frac{8}{11} $$

$$ \displaystyle \frac{6}{11} $$

Solution

Question 2

$$P,Q$$

$$Q,R$$

$$P,R$$

None of these

Solution

Question 3

$$\displaystyle \frac{5}{12}$$

$$\displaystyle \frac{3}{8}$$

$$\displaystyle \frac{4}{5}$$

$$\displaystyle \frac{5}{4}$$

Solution

Question 4

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 5$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 2$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 4$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 8$$

Solution

Question 5

$$\displaystyle \frac{5}{12}$$

$$\displaystyle \frac{6}{12}$$

$$\displaystyle \frac{7}{12}$$

$$\displaystyle \frac{11}{12}$$

Solution

Question 6

$$n(S) = 10, n(P) = 7, n(Q) = 1, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 3, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 2, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$

Solution

Question 7

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

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

$$\displaystyle \frac{1}{2}$$

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

Solution

Question 8

0.3

0.4

0.5

0.6

Solution

Question 9

$$n(S) = 9, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 4, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 6, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 7, n(P) =3, n(Q) = 6, n(R) = 6$$

Solution

Question 10

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

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

$$\displaystyle \frac{1}{2}$$

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

Solution

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