Probability Test

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Probability Test - 55

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Question 1
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Given that $$A, B$$ and $$C$$ are events such that $$P(A)=P(B)=P(C)=1/5, P(A\cap B)=P(B\cap C)=0$$ and $$P(A\cap C)=1/10$$. The probability that at least one of the events $$A, B$$ or $$C$$ occurs is

A
$$1/2$$

B
$$2/3$$

C
$$2/5$$

D
$$3/5$$

Solution

We have $$A\cap B\cap C\underline {\cap} A\cap B$$
$$\Rightarrow P(A\cap B\cap C)\leq P(A\cap B)=0$$
Since, $$P(A\cap B\cap C)\geq 0$$, we get $$P(A\cap B\cap C)=0$$.
Now, $$P$$ (at least one of $$A, B, C$$)
$$\displaystyle =P(A\cap B\cap C)=P(A)+P(B)+P(C)-P(B\cap C)-P(C\cap A)-P(A\cap B)+P(A\cap B\cap C)$$
$$\displaystyle \frac {1}{5}+\frac {1}{5}+\frac {1}{5}-0-\frac {1}{10}-0+0=\frac {1}{2}$$.
Question 2
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For the three events $$A, B$$ and $$C, P$$ (exactly one of the events $$A$$ or $$B$$ occurs) $$=$$ P(exactly one of the events $$B$$ or $$C $$ occurs) $$=$$ P(exactly one of the events $$C$$ or $$A$$ occurs) $$=$$ $$p$$ and P(all the three events occur simultaneously) $$=p^2$$, where $$0 < p < 1/2$$. Then the probability of at least one of the three events $$A, B$$ and $$C$$ occurring is

A
$$\dfrac {3p+2p^2}{2}$$

B
$$\dfrac {p+3p^2}{4}$$

C
$$\dfrac {p+3p^2}{2}$$

D
$$\dfrac {3p+2p^2}{4}$$

Solution

We know that
P(exactly one of $$ A$$ or $$B$$ occurs)
$$=P(A)+P(B)-2P(A\cap B)$$
Therefore, $$P(A)+P(B)-2P(A\cap B)=p$$ -----(1)
Similarly, $$P(B)+P(C)-2P(B\cap C)=p$$ -----(2)
and $$P(C)+P(A)-2P(C\cap A)=p$$ -----(3)
Adding (1), (2) and (3) we get
$$2[P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)]=3p$$
$$\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)=3p/2$$ (4)
We are also given that $$P(A\cap B\cap C)=p^2$$ (5)
Now, P (at least one of $$A, B$$ and $$C$$)
$$=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$$
$$=\dfrac {3p}{2}+p^2=\dfrac {3p+2p^2}{2}$$ [using (4) and (5)]
Question 3
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A bag contains $$a $$ white and $$b$$ black balls. Two players, $$A$$ and $$B$$ alternately draw a ball from the bag, replacing the ball each time after the draw till on of them draws a white ball and win the game. $$A$$ begins the game. If the probability of $$A$$ winning the game is three times that of $$B$$, the ratio $$a:b$$ is

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
None of these

Solution

Let $$W$$ denote the event of drawing a white ball at any draw and $$B$$ that for a black ball.
Then $$\displaystyle P(W)=\frac {a}{a+b}$$ and $$P(B)=\dfrac {b}{a+b}$$
$$P $$($$A $$ wins the game) $$=$$ $$P$$($$W$$ or $$BBW$$ or $$BBBBW$$ or$$ ....$$)
$$=P(W)+P(BBW)+P(BBBBW)+.....$$
$$=P(W)+P(B)P(B)P(W)+P(B)P(B)P(B)P(B)P(W)+....$$
$$=P(W)+P(W)\cdot P(B)^2+P(W)\cdot P(B)^4+.....$$
$$\displaystyle=\frac {P(W)}{1-P(B)^2}=\frac {a(a+b)}{a^2+2ab}=\frac {a+b}{a+2b}$$
Also $$P$$($$B$$ wins the game)$$=\displaystyle1-\frac {a+b}{a+2b}=\frac {b}{a+2b}$$
According to the given conditions,
$$\displaystyle\Rightarrow \frac {a+b}{a+2b}=3\cdot \frac {b}{a+2b}\Rightarrow a=2b\Rightarrow a:b=2:1$$
Question 4
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For three events $$A, B$$ and $$C, P$$ (exactly one of the events $$A$$ occur) $$=$$ P (exactly one of the events B and C occur) $$=$$ P (exactly one of the events C or A occurs) $$=$$ p and P (all the three events occur simultaneously) $$=p^2$$, where $$0 < p < 1/2$$. If the probability of at least one of the three events $$A, B$$ and $$C$$ occurs is $$11/18$$, the value of $$p$$ is

A
$$1/6$$

B
$$1/4$$

C
$$1/5$$

D
$$1/3$$

Solution

P (exactly one of A or B occurs)
$$=P(A)+P(B)-2P(A\cap B)$$
Therefore, $$P(A)+P(B)-2P(A\cap B)=p$$ (1)
Similarly, $$P(B)+P(C)-2P(B\cap C)=p$$ (2)
and $$P(C)+P(A)-2P(C\cap A)=p$$ (3)
Adding (1), (2) and (3) and dividing by 2
$$\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)=3p/2$$ (4)
Now, P (at least one of A, B and C)
$$\displaystyle =\frac {3p}{2}+p^2=\frac {3p+2p^2}{2} =\frac{11}{18}$$ (given)
$$\Rightarrow \displaystyle 2p^2+3p-\frac{11}{9}=0\therefore p = \frac{-3+\sqrt{9+\frac{88}{9}}}{4}=\frac{1}{3}$$
Question 5
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If $$P(A)=0.7, P(B)=0.55, P(C)=0.5, P(A\cap B)=x, P(A\cap C)=0.45, P(B\cap C)=0.3$$ and $$P(A\cap B\cap C)=0.2$$, then

A
$$0.2\leq x\leq 0.45$$

B
$$0.2\leq x\leq 0.5$$

C
$$0.2\leq x\leq 0.65$$

D
$$0.2\leq x\leq 1$$

Solution

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

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

$$ =>P(A\cap B)=0.7+0.55+0.5-P(A\cup B\cup C)-0.45-0.3+0.2=1.2-P(A\cup B\cup C)$$

Maximum $$P(A\cup B\cup C)$$ is 1, hence, minimum $$P(A\cap B)$$ = $$1.2-1=0.2$$

Maximum $$P(A\cap B)$$ could have been the minimum of $$P(A)$$ and $$P(B)$$ i.e. $$0.55$$,

however, that would mean that $$1.2-P(A\cup B\cup C)=0.55 => P(A\cup B\cup C)=0.65$$, i.e. less than A: This is not possible

hence, Maximum $$P(A\cap B)$$ could be $$1.2-0.7=0.5$$.

$$\therefore\ 0.2 \le P(A\cap B) \le 0.5\Rightarrow 0.2 \le x \le 0.5$$
Question 6
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Twelve players $${ S }_{ 1 },{ S }_{ 2 },...,{ S }_{ 12 }$$ play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that at least one of $${S}_{1}$$ and $${S}_{2}$$ is among the six winners is:

A
$$\displaystyle \frac { 31 }{ 33 } $$

B
$$\displaystyle \frac { 32 }{ 33 } $$

C
$$\displaystyle \frac { 10 }{ 11 } $$

D
None of these

Solution
Question 7
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Directions For Questions

In a class, there are $$100$$ students out of which $$45$$ study mathematics, $$48$$ study physics, $$40$$ study chemistry, $$12$$ study both mathematics & physics, $$11$$ study both physics & chemistry, $$15$$ study both mathematics & chemistry and $$5$$ study all three subjects. A student is selected at random, then find the probability that the selected student studies

...view full instructions

Only one subject

A
$$0.72$$

B
$$0.36$$

C
$$0.48$$

D
$$0.24$$

Solution

Given, In a class $$100$$ students out of which $$45$$ study mathematics, $$48$$ study physics, $$40$$ study chemistry, $$12$$ study both maths and physics, $$11$$ study both physics and chemistry, $$15$$ study both mathematics and chemistry and $$5$$ study all the subjects.
$$\Rightarrow n(\mu)=100,n(M)=45,n(P)=48,n(C)=40,n(M \cap P)=12,n(P \cap C)=11$$
$$n(C \cap M)=15,n(M \cap P \cap C)=5$$
Student studies only one subject,
$$n(X)=(n(M)-n(M \cap P)-n(C \cap M)n(M \cap P \cap C))$$
$$+(n(C)-n(P \cap C)-n(C \cap M)+n(M \cap P \cap C))+(n(C)-n(P \cap C)-n(C \cap M)+n(M \cap P \cap C))$$
$$=(45-12-15+5)+(40-11-15+5)+(48-11-12+5)$$$$=23+19+30=72$$
The required probability$$=\displaystyle\frac{n(X)}{n(\mu)}=\frac{72}{100}=0.72$$
Question 8
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Three shots are fired at a target in succession. The probabilities of a hit in the first shot is $$\displaystyle \frac {1}{2}$$, in the second $$\displaystyle \frac {2}{3}$$ and in the third shot is $$\displaystyle \frac {3}{4}$$, In case of exactly one hit, the probability of destroying the target is $$\displaystyle \frac {1}{3}$$ and in the case of exactly two hits $$\displaystyle \frac {7}{11}$$ an in the case of three hits is $$1.0$$. Find the probability of destroying the target in three shots

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

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

C
$$\displaystyle \frac {5}{8}$$

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

Solution

$$A$$ : Target hit in 1st shot
$$B$$ : Target hit in 2nd shot
$$C$$ : Target hit in 3rd shot
$$E_1$$: destroyed in exactly one shot
$$E_2$$: destroyed in exactly two shot
$$E_3$$: destroyed in exactly three shot
$$P(E_1) =P(E_1A\overline B\overline C \cup E_1\overline A\overline BC \cup E_1\overline AB\overline C)$$
$$=\dfrac{1}{3}\left [\dfrac {1}{2}.\dfrac {1}{3}.\dfrac {1}{4}+\dfrac {1}{2}.\dfrac{1}{3}.\dfrac {3}{4}+\dfrac {1}{2}.\dfrac {2}{3}.\dfrac {1}{4}\right ]=\dfrac{1+3+2}{3.24}=\dfrac {1}{12}$$
$$P(E_2)= P(E_2\overline ABC \cup E_2AB\overline C \cup E_3A\overline BC)$$
$$=\dfrac{7}{11}\left [\dfrac {1}{2}.\dfrac {2}{3}.\dfrac {3}{4}+\dfrac {1}{2}.\dfrac{2}{3}.\dfrac {1}{4}+\dfrac {1}{2}.\dfrac {1}{3}.\dfrac {3}{4}\right]=\dfrac {7.11}{11.24}=\dfrac {7}{24}$$
$$P(E_3)=P(E_3ABC)=1.\dfrac {1}{2}.\dfrac {2}{3}.\dfrac {3}{4}=\dfrac {1}{4}$$
$$P(E_1\cup E_2\cup E_3)=P(E_1)+P(E_2)+P(E_3)$$
$$=\dfrac {1}{12}+\dfrac {7}{24}+\dfrac {1}{4}=\dfrac {2+7+6}{24}=\dfrac {15}{24}=\dfrac {5}{8}$$
Question 9
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A survey of people in a given region showed that $$20$$% were smokers. The probability of death due to lung cancer, given that a person smoked, was $$10$$ times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is $$0.006$$, what is the probability of death due to lung cancer given that a person is a smoker?

A
$$1/140$$

B
$$1/70$$

C
$$3/140$$

D
$$1/10$$

Solution

Let L be the event of lung cancer, S be the event of smoker and S' be the event of a non-smoker.
Given:
$$P( L/S) = 10 \times P( L/S') $$
$$P( L ) = 0.006 $$
$$P(S)=0.2\\\implies P(S')=1-P(S)=0.8$$
To find:
The probability of death due to lung cancer given that a person is a smoker, P(L/S)
We know, using the law of total probability that
$$P(L)=P(L/S)\times P(S) + P(L/S')\times P(S')$$
Substituting the values, we get
$$0.006 = P(L/S)\times 0.2 + P(L/S')\times 0.8 $$
$$0.006 = 0.2 \times P(L/S) +\dfrac 1{10}\times P(L/S)\times 0.8 $$
$$0.006 = P(L/S)\times\left ( 0.2 +\dfrac 1{10}\times 0.8\right) $$
$$0.006 = P( L/S)\times 0.28 $$
$$0.02142857 = P( L/S)$$
or, The probability of death due to lung cancer given that a person is a smoker = $$\dfrac 3{140}$$
Question 10
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A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

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

B
$$\dfrac{1}{14}$$

C
$$\dfrac{2}{63}$$

D
$$\dfrac{2}{14}$$

Solution

Total Blue Pens: $$4$$
Total Red Pens: $$2$$
Total Black Pens: $$3$$
Total Pens: $$4+2+3 = 9$$
Probability of drawing $$2$$ blue pens $$=\dfrac{\ ^4C_2}{\ ^9C_2} = \dfrac{4\times3}{9\times8} = \dfrac16$$
After this, the pens are not replaced, which reduces the number of pens in the pack to $$7$$.
So, the probability of drawing $$1$$ black pen from a pack of $$7$$ pens would be $$\dfrac{\ ^3C_1}{\ ^7C_1} = \dfrac37$$
Probability of drawing $$2$$ blue pens and $$1$$ black pen $$=\dfrac16\times \dfrac{3}{7} = \dfrac{1}{14}$$.

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Probability Test - 55

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Probability Test - 55

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SHARING IS CARING

Question 1

$$1/2$$

$$2/3$$

$$2/5$$

$$3/5$$

Solution

Question 2

$$\dfrac {3p+2p^2}{2}$$

$$\dfrac {p+3p^2}{4}$$

$$\dfrac {p+3p^2}{2}$$

$$\dfrac {3p+2p^2}{4}$$

Solution

Question 3

$$1:1$$

$$1:2$$

$$2:1$$

None of these

Solution

Question 4

$$1/6$$

$$1/4$$

$$1/5$$

$$1/3$$

Solution

Question 5

$$0.2\leq x\leq 0.45$$

$$0.2\leq x\leq 0.5$$

$$0.2\leq x\leq 0.65$$

$$0.2\leq x\leq 1$$

Solution

Question 6

$$\displaystyle \frac { 31 }{ 33 } $$

$$\displaystyle \frac { 32 }{ 33 } $$

$$\displaystyle \frac { 10 }{ 11 } $$

None of these

Solution

Question 7

$$0.72$$

$$0.36$$

$$0.48$$

$$0.24$$

Solution

Question 8

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

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

$$\displaystyle \frac {5}{8}$$

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

Solution

Question 9

$$1/140$$

$$1/70$$

$$3/140$$

$$1/10$$

Solution

Question 10

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

$$\dfrac{1}{14}$$

$$\dfrac{2}{63}$$

$$\dfrac{2}{14}$$

Solution

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