Probability Test

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Probability Test - 44

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Weekly Quiz Competition

Question 1
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If $$A$$ and $$B$$ are two events such that $$P(A \cup B) = \dfrac {3}{4}. P(A\cap B) = \dfrac {1}{4}$$ and $$P(\overline {A}) = \dfrac {2}{3}$$, then what is $$P(B)$$ equal to?

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

B
$$\dfrac{2}{3}$$

C
$$\dfrac{1}{8}$$

D
$$\dfrac{2}{9}$$

Solution

Given $$P(A\cup B)=\cfrac { 3 }{ 4 } , P(A\cap B)=\cfrac { 1 }{ 4 }$$
$$P(A)=1-P(\overline{A})=\dfrac{1}{3}$$
$$\Longrightarrow P(A)+P(B)-P(A\cap B)=P(A\cup B)=\dfrac { 3 }{ 4 } \\ \Longrightarrow \dfrac { 1 }{ 3 } +P(B)-\cfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 } \\ \Longrightarrow \dfrac { 1 }{ 3 } +P(B)=1\\ \Longrightarrow P(B)=\dfrac { 2 }{ 3 } $$
Question 2
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Simone and her three friends were deciding how to pick the song they will sing for their school's talent show. They decide to roll a number cube.
The person with the lowest number chooses the song. If her friends rolled a 6, 5, and 2, what is the probability that Simone will get to choose the song?

A
$$\dfrac{1}{6}$$

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

C
$$0$$

D
$$1$$

Solution

The possible outcomes of rolling a number cube are $$1, 2, 3, 4, 5, 6 $$.

In order for Simone to be able to choose the song, she will need to roll a $$1$$.

Let $$P(A)$$ be the probability that Simone chooses the song.

$$P(A) = \dfrac{number\:of\:favorable\:outcomes}{number\:of\:possible\:outcome}$$
$$=\dfrac{1}{6}$$ (there are $$6$$ possible outcomes, and $$1$$ of them is favorable)

The probability that Simone will choose the song is $$\dfrac{1}{6}$$.
Question 3
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There are $$n$$ persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together, is

A
$$\cfrac { 2 }{ n } $$

B
$$1-\cfrac { 2 }{ n } $$

C
$$\cfrac { n(n-1) }{ (n+1)(n+2) } $$

D
None of these

Solution

Total persons= $$n$$

Total number of wage to select two at a time = $$^{ n }{ C }_{ 2 }$$
No. of ways to select two together person= $$n-1$$

Probability of two together person,
= $$\cfrac { n-1 }{ ^{ n }{ C }_{ 2 } } $$
= $$\cfrac { n-1 }{ { n! }/{ (n-2)!2! } } $$

= $$\cfrac { (n-1)(n-2)!2 }{ n(n-1)(n-2)! } $$

= $$1-\cfrac { 2 }{ n } $$
Question 4
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The probability that a man will live $$10$$ more years, is $$\dfrac { 1 }{ 4 }$$ and the probability that his wife will live $$10$$ more years, is $$\dfrac { 1 }{ 3 }$$. Then, what is the probability that neither will be alive in $$10$$ years?

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

B
$$\dfrac { 3 }{ 7 } $$

C
$$\dfrac { 2 }{ 3 } $$

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

Solution

let M be the event that the man will live 10 years more
let W be the event that the wife will live 10 years more

Given, $$P\left( \ { M } \right) =\dfrac { 1 }{ 4 } \Rightarrow P\left( \bar { M } \right) =1-\dfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 } $$
and $$P\left( \ { W } \right) =\dfrac { 1 }{ 3 } \Rightarrow P\left( \bar { W } \right) =1-\dfrac { 1 }{ 3 } =\dfrac { 2 }{ 3 } $$
Here, both events are independent, so the required probability
$$=P\left( \bar { W } \cap \bar { M } \right) $$
$$=P\left( \bar { W } \right) \cdot P\left( \bar { M } \right) $$
$$=\dfrac { 3 }{ 4 } \times \dfrac { 2 }{ 3 } $$
$$=\dfrac { 1 }{ 2 } $$
Question 5
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A coin is tossed three times. Consider the following events:
A: No head appears
B: Exactly one head appears
C: At least two heads appear
Which one of the following is correct?

A
$$(A \cup B) \cap (A \cup C) = B \cup C$$

B
$$(A \cap B') \cup (A \cap C') = B' \cup C'$$

C
$$A \cap B (B' \cup C') = A \cup B \cup C$$

D
$$A \cap (B' \cup C') = B' \cap C'$$

Solution

$$A=\{(T,T,T)\}$$, $$B=\{(H,T,T),(T,H,T),(T,T,H)\}$$ and $$C=\{(H,H,T),(T,H,H),(H,T,H),(H,H,H)\}$$
$$\therefore (A\cup B)\cap (A\cup C)$$
$$=\{(T,T,T),(H,T,T),(T,H,T),(T,T,H)\}\cap \{(T,T,T),(H,H,T),(T,H,H),(H,T,H),(H,H,H)\}$$
$$=\{(T,T,T)\}=A$$
$$\therefore (A\cap B' )\cup (A\cap C' )$$
$$=\{(H,H,T),(T,H,H),(H,T,H),(H,H,H)\}\cup \{(T,T,T),(H,T,T),(T,H,T),(T,T,H)\}=B'\cup C'$$
$$\therefore A\cap B\cap (B' \cup C' )$$
$$=\{(T,T,T)\}\cap \{(H,T,T),(T,H,T),(T,T,H)\} \cap \{(T,T,T),(H,T,T)$$
$$,(T,H,T),(T,T,H),(H,H,T),(T,H,H),(H,T,H),(H,H,H)\}$$
$$=\phi$$
$$\therefore A\cap (B' \cup C' )$$
$$=\{(T,T,T)\}\cap \{(H,H,T),(T,H,H),(H,T,H),(H,H,H),(T,T,T),(H,T,T),(T,H,T),(T,T,H)\}$$
$$= \{(T,T,T)\}=A$$
Question 6
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From $$50$$ students taking examinations in Mathematics, Physics and Chemistry, $$37$$ passed in Mathematics, $$24$$ in Physics and $$43$$ in Chemistry. Atmost $$19$$ passed in Mathematics and Physics, atmost $$29$$ passed in Mathematics and Chemistry and atmost $$20$$ passed in Physics and Chemistry. The largest possible number that could have passed all three examination, is?

A
$$11$$

B
$$12$$

C
$$13$$

D
$$14$$

Solution

Let M, P and C be the sets of students taking examinations in Mathematics, Physics and Chemistry, respectively.
We have,

$$n(M \cup P \cup C)=50, n(M)=37, n(P)=24$$,
$$n(C)=43$$
$$n(M\cap P) \leq 19, n(M \cap C) \leq 29, n(P \cap C)\leq 20$$

Now, $$n(M \cup P\cup C)=n(M)+n(P)+n(C)-n(M\cap P)-n(M\cap C) - n(P \cap C)+n(M\cap P\cap C)$$

$$\Rightarrow 50=37+24+43-\{n(M\cap P)+n(M \cap C)+n(P \cap C)\}+n(M\cap P \cap C)$$

$$\Rightarrow n(M\cap P)+n(M\cap C)+n(P \cap C)=n(M\cap P\cap C)+54$$ .......(i)

Now, $$n(M\cap P)+n(M\cap C)+n(P\cap C)\leq 19+29+20$$ [using Eq. (i)]
$$\Rightarrow n(M\cap P\cap C)+54\leq 68$$

$$\Rightarrow n(M\cap P\cap C)\leq 14$$.
Question 7
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A purse contains $$4$$ copper coins and $$3$$ silver coins. A second purse contains $$6$$ copper coins and $$4$$ silver coins. A purse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?

A
$$\dfrac{41}{70}$$

B
$$\dfrac{31}{70}$$

C
$$\dfrac{27}{70}$$

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

Solution

$${\textbf{Step - 1: Find separate probabilities of every event}}{\text{.}}$$
$${\text{Let }}{E_1}{\text{ be the event of drawing a copper coin from the first purse}}{\text{.}}$$
$${\text{We can write }}P({E_1}) = \dfrac{{{\text{number of copper coins in purse 1}}}}{{{\text{total number of coins in purse 1}}}}$$
$$P({E_1}) = \dfrac{4}{7}$$
$${\text{Let }}{E_2}{\text{ be the event of drawing a copper coin from the second purse}}{\text{.}}$$
$${\text{We can write }}P({E_2}) = \dfrac{{{\text{number of copper coins in purse 2}}}}{{{\text{total number of coins in purse 2}}}}$$
$$P({E_2}) = \dfrac{6}{10}$$
$${\text{Also, let }}{E_3}{\text{ be an event of choosing a purse between the two,}}$$
$${\text{As both the purses have equal probability of getting chosen }}P({E_3}) = \dfrac{1}{2}$$
$${\textbf{Step - 2: Find the probability of the drawn coin being a copper coin}}$$
$${\text{Let }}E{\text{ be the event of drawing a copper coin, we can write}}$$
$$P(E) = ({\text{choosing the first purse)}} \times ({\text{probability of choosing a copper coin from purse 1)}}$$
$${\text{ + (probability of choosing purse 2)}} \times ({\text{probability of choosing copper coin from purse 2)}}$$
$$P(E) = P({E_3}) \times P({E_1}) + P({E_3}) \times P({E_2})$$
$$P(E) = \dfrac{1}{2} \times \dfrac{4}{7}{\text{ + }}\dfrac{1}{2} \times \dfrac{6}{10}$$
$$P(E) = \dfrac{2}{7} + \dfrac{3}{10} = \dfrac{{20 + 21}}{{70}} = \dfrac{{41}}{{70}}$$
$$\textbf{Hence option A is correct}$$
Question 8
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Two events A and B are such that $$P($$not $$B)$$ $$= 0.8$$, $$P(A\cup B)=0.5$$ and $$P(A\setminus B)=0.4$$ Then $$P(A)$$ is equal to

A
$$0.28$$

B
$$0.32$$

C
$$0.38$$

D
None of the above

Solution

Solution:
Given: $$P(\bar B)=0.8\therefore P(B)=1-P(\bar B)=1-0.8=0.2$$
$$P(A\cup B)=0.5$$ and $$P(A\setminus B)=0.4$$
$$\therefore P(A\setminus B)=\cfrac{P(A\cap B)}{P(B)}$$
or, $$0.4=\cfrac{P(A\cap B)}{0.2}$$
or, $$P(A\cap B)=0.08$$
Hence, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
or, $$0.5=P(A)+0.2-0.08$$
or, $$P(A)=0.38$$
Hence, C is the correct option.
Question 9
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Probability of solving a particular question by person $$A$$ is $$\dfrac { 1 }{ 3 }$$ and probability of solving that question by person $$B$$ is $$\dfrac { 2 }{ 5 } $$. What is the probability of solving that question by atleast one of them?

A
$$\dfrac { 3 }{ 5 } $$

B
$$\dfrac { 7 }{ 9 } $$

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

D
$$\dfrac { 2 }{ 3 } $$

Solution

Given, $$P\left( A \right) =\dfrac { 1 }{ 3 }$$ and $$ P\left( B \right) =\dfrac { 2 }{ 5 }$$

$$ \therefore P\left( \bar { A } \right) =1-P\left( A \right) =1-\dfrac { 1 }{ 3 } =\dfrac { 2 }{ 3 }$$
and $$ P\left( \bar { B } \right) =1-P\left( B \right) =1-\dfrac { 2 }{ 5 } =\dfrac { 3 }{ 5 } $$

Now, $$P\left( atleast\quad one\quad of\quad them \right)$$

$$ =P\left( A\cap \bar { B } \right) +P\left( \bar { A } \cap B \right) +P\left( A\cap B \right) $$
$$=P\left( A \right) \cdot P\left( \bar { B } \right) +P\left( \bar { A } \right) \cdot P\left( B \right) +P\left( A \right) \cdot P\left( B \right) $$
[$$\because A$$ and $$B$$ are independent events]

$$=\dfrac { 1 }{ 3 } \cdot \dfrac { 3 }{ 5 } +\dfrac { 2 }{ 3 } \cdot \dfrac { 2 }{ 5 } +\dfrac { 1 }{ 3 } \cdot \dfrac { 2 }{ 5 } $$

$$=\dfrac { 3 }{ 15 } +\dfrac { 4 }{ 15 } +\dfrac { 2 }{ 15 } =\dfrac { 9 }{ 15 } =\dfrac { 3 }{ 5 } $$.
Question 10
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A box contains $$6$$ green balls, $$4$$ blue balls and $$5$$ yellow balls. A ball is drawn at random. Find the probability of
(a) Getting a yellow ball.
(b) Not getting a green ball.

A
$$\dfrac{1}{5},\dfrac{1}{3}$$

B
$$\dfrac{4}{15}, \dfrac{3}{15}$$

C
$$\dfrac{1}{3}, \dfrac{3}{5}$$

D
$$\dfrac{2}{3}, \dfrac{1}{15}$$

Solution

A box contains $$6$$ green balls, $$4$$ blue balls, $$5$$ yellow balls.

Total number of balls $$n(S)=6+4+5=15$$

$$(a)$$

Let $$A$$ be the probability of getting yellow ball.

$$n(A)=5$$

Thus the probability of getting yellow ball is $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{5}{15}=\dfrac{1}{3}$$.

$$(b)$$

Let $$B$$ be the probability of not getting green ball. That is, probability of getting blue and yellow balls.

$$n(B)=4+5=9$$

Thus the probability of getting yellow ball is $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{9}{15}=\dfrac{3}{5}$$.

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Probability Test - 44

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Probability Test - 44

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SHARING IS CARING

Question 1

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

$$\dfrac{2}{3}$$

$$\dfrac{1}{8}$$

$$\dfrac{2}{9}$$

Solution

Question 2

$$\dfrac{1}{6}$$

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

$$0$$

$$1$$

Solution

Question 3

$$\cfrac { 2 }{ n } $$

$$1-\cfrac { 2 }{ n } $$

$$\cfrac { n(n-1) }{ (n+1)(n+2) } $$

None of these

Solution

Question 4

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

$$\dfrac { 3 }{ 7 } $$

$$\dfrac { 2 }{ 3 } $$

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

Solution

Question 5

$$(A \cup B) \cap (A \cup C) = B \cup C$$

$$(A \cap B') \cup (A \cap C') = B' \cup C'$$

$$A \cap B (B' \cup C') = A \cup B \cup C$$

$$A \cap (B' \cup C') = B' \cap C'$$

Solution

Question 6

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Question 7

$$\dfrac{41}{70}$$

$$\dfrac{31}{70}$$

$$\dfrac{27}{70}$$

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

Solution

Question 8

$$0.28$$

$$0.32$$

$$0.38$$

None of the above

Solution

Question 9

$$\dfrac { 3 }{ 5 } $$

$$\dfrac { 7 }{ 9 } $$

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

$$\dfrac { 2 }{ 3 } $$

Solution

Question 10

$$\dfrac{1}{5},\dfrac{1}{3}$$

$$\dfrac{4}{15}, \dfrac{3}{15}$$

$$\dfrac{1}{3}, \dfrac{3}{5}$$

$$\dfrac{2}{3}, \dfrac{1}{15}$$

Solution

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