Probability Test

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Probability Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

A medicine is known to be 75% effective to cure a patient. If the medicine is given to 5 patients, what is the probability that at least one patient is curved by this medicine?

A
$$\dfrac{1}{1024}$$

B
$$\dfrac{243}{1024}$$

C
$$\dfrac{1023}{1024}$$

D
$$\dfrac{781}{1024}$$

Solution

Let p be the probability that a patient is cured by the medicine
$$p=75\%= \dfrac{3}{4} $$
Probability that a patient is not cured by medicine $$=q=1-p=\dfrac{1}{4}$$
Let P(X) be the probability that X patients are cured by the medicine
Probability that at least one patient is cured $$=P(X\geq 1)$$
$$=1-P(X=0)$$
$$=1-^5C_0(p)^0(q)^5$$

$$=1-\left(\dfrac{1}{4}\right)^5$$

$$=\dfrac{1023}{1024}$$

Answer is option (C)
Question 2
1 / -0

A machine has three parts, A, B and C, whose chances of being defective are 0.02, 0.10 and 0.05 respectively. The machine stops working if any one of the arts becomes defective. What is the probability that the machine will not stop working?

A
$$0.06$$

B
$$0.16$$

C
$$0.84$$

D
$$0.94$$

Solution

Let $$P(A), P(B),P(C)$$ be the probability of parts A,B and C working properly.
$$\therefore P(A)=1-0.02=0.98$$
$$\therefore P(B)=1-0.1=0.9$$
$$\therefore P(C)=1-0.05=0.95$$

Probaility that machine will not stop working $$=$$ Probaility that all parts are working properly
$$=P(A\cap B\cap C )$$
$$=P(A)P(B)P(C)$$ ...( A,B and C are independent events)
$$=(0.98)(0.9)(0.95)$$
$$=0.8379=0.84$$

Answer is option (C)
Question 3
1 / -0

There is $$25$$% chance that it rains on any particular day. What is the probability that there is at least one rainy day within a period of $$7$$ days?

A
$$1 - \left( \dfrac{1}{4} \right )^7$$

B
$$\left( \dfrac{1}{4} \right )^7$$

C
$$\left( \dfrac{3}{4} \right )^7$$

D
$$1- \left( \dfrac{3}{4} \right )^7$$

Solution

Let p be the probability that it rains on a particular day.
$$ p=25\%=\dfrac{1}{4}$$

Probability that there is at least 1 rainy day within a period of 7 days
$$=1-$$ Probability that there is no rainy day within a period of 7 days
$$=1-^7C_{0}p^0(1-p)^7$$

$$=1-\left(1-\dfrac{1}{4}\right)^7$$

$$=1-\left(\dfrac{3}{4}\right)^7$$
Answer is option (D)
Question 4
1 / -0

Three independent events, $$A_1, A_2$$ and $$A_3$$ occur with probabilities $$P(A_i) = \dfrac{1}{1 + i}, i = 1, 2, 3$$. What is the probability that at least one of the three events occurs?

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

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

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

D
$$\dfrac{1}{24}$$

Solution

$$P(A_i)=\dfrac{1}{1+i}$$

$$\therefore P(A_1)=\dfrac{1}{1+1}=\dfrac{1}{2}, P(A_2)=\dfrac{1}{3}, P(A_3)=\dfrac{1}{4}$$

Probability that at least one event occurs = 1 - Probability that none of $$A_1,A_2$$ or $$A_3$$ occur
$$=1-P(\overline{A_1}\cap \overline{A_2}\cap \overline{A_3})$$

$$=1-P(\overline{A_1})P(\overline{A_2})P(\overline{A_3})$$ ...($$A_1,A_2$$ and $$A_3$$ are independent events)

$$=1-\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)$$

$$=1-\left(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\right)$$

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

Answer is option (C)
Question 5
1 / -0

In a series of 3 one-day cricket matches between teams A and B of a college, the probability of team A winning or drawing are 1/3 and 1/6 respectively. If a win, loss or draw gives 2, 0 and 1 point respectively, then what is the probability that team A will score 5 points in the series?

A
$$\dfrac{17}{18}$$

B
$$\dfrac{11}{12}$$

C
$$\dfrac{1}{12}$$

D
$$\dfrac{1}{18}$$

Solution

To score 5 points, team A need to win 2 matches and draw 1 match.
Let W be the event that team A wins the game.
Let D be the event that team A draws the game.
To score 5 points, possible outcomes of 3 matches are $$WWD,WDW,DWW$$

Probability to score 5 points $$=P(WWD)+P(WDW)+P(DWW)$$

$$=P(W)P(W)P(D)+P(W)P(D)P(W)+P(D)P(W)P(W) $$ ...(W and D are independent events)

$$=\left(\dfrac{1}{3}\times \dfrac{1}{3}\times \dfrac{1}{6}\right)+\left(\dfrac{1}{3}\times \dfrac{1}{6}\times \dfrac{1}{3}\right)+\left(\dfrac{1}{6}\times \dfrac{1}{3}\times\dfrac{1}{3}\right)$$

$$=\dfrac{1}{18}$$
Question 6
1 / -0

A salesman has a $$70\%$$ chance to sell a product to any customer. The behaviour of successive customers is independent. If two customers A and B enter, what is the probability that the salesman will sell the product to customer A or B?

A
$$0.98$$

B
$$0.91$$

C
$$0.70$$

D
$$0.49$$

Solution

Let $$P(A)$$ and $$P(B)$$ be the probability that salesman sells the product to costumer $$A$$ and $$B$$ respectively
$$\therefore P(A)=P(B)=70\%=0.7$$

Probability that the salesman will sell the product to customer $$A$$ or $$B=P(A\cap \overline{B})+P(\overline{B}\cap A )+P(A\cap B)$$
$$=P(A)P(\overline{B})+P(\overline{A})P(B)+P(A)(B)$$ ...( $$P(A)$$ and $$P(B)$$ are independent)
$$=(0.7)(1-0.7)+(1-0.7)(0.7)+(0.7)(0.7)$$
$$=0.21+0.21+0.49$$
$$=0.91$$
Answer is option (B).
Question 7
1 / -0

For two events, A and B, it is given that $$P(A) = \dfrac{3}{5}, P(B) = \dfrac{3}{10} $$ and $$P(A|B) = \dfrac{2}{3}$$. If $$\overline A$$ and $$\overline B$$ are the complementary events of A and B, then what is $$P(\overline A | \overline B)$$ equal to?

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

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

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

D
$$\dfrac{4}{7}$$

Solution

$$P(A/B)=\dfrac{P(A\cap B) }{P(B)}$$
$$\therefore \dfrac{2}{3}=\dfrac{P(A\cap B) }{\dfrac{3}{10}}$$
$$\therefore P(A\cap B) =\dfrac{1}{5}$$
$$\therefore P(A)+P(B)-P(A\cup B )=\dfrac{1}{5}$$
$$\therefore \dfrac{3}{5}+\dfrac{3}{10}-P(A\cup B )=\dfrac{1}{5}$$
$$\therefore P(A\cup B )=\dfrac{7}{10}$$

$$P(\overline{A}/\overline{B})=\dfrac{P(\overline{A}\cap \overline{B}) }{P(\overline{B})}$$
$$=\dfrac{1-P(A\cup B) }{1-P(B)}$$

$$=\dfrac{1-\dfrac{7}{10}}{1-\dfrac{3}{10}}$$
$$=\dfrac{3}{7}$$

Answer is option (A)
Question 8
1 / -0

Three candidates solve a question. Odds in favour of the correct answer are $$5 : 2, 4 : 3$$ and $$3 : 4$$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?

A
$$\dfrac{209}{343}$$

B
$$\dfrac{134}{343}$$

C
$$\dfrac{149}{343}$$

D
$$\dfrac{60}{343}$$

Solution

Let A,B,C be the three students.
Let P(A), P(B), P(C) be the probabilities that students A,B and C answer the question correctly.
$$P(A)=\dfrac{5}{5+2}=\dfrac{5}{7}$$

$$P(B)=\dfrac{4}{4+3}=\dfrac{4}{7}$$

$$P(C)=\dfrac{3}{3+4}=\dfrac{3}{7}$$

Probability that atleast 2 of them solve the question correctly
$$=P(A\cap B\cap \overline{C})+P(A\cap \overline{B} \cap C)+P(\overline{A} \cap B\cap C)+P(A\cap B\cap C)$$

$$=P(A)P(B)P(\overline{C})+P(A)P(\overline{B})P(C)+P(\overline{A})P(B)P(C)+P(A)P(B)P(C)$$

$$=(\dfrac{5}{7})(\dfrac{4}{7})(1-\dfrac{3}{7})+(\dfrac{5}{7})(1-\dfrac{4}{7})(\dfrac{3}{7})+(1-\dfrac{5}{7})(\dfrac{4}{7})(\dfrac{3}{7})+(\dfrac{5}{7})(\dfrac{4}{7})(\dfrac{3}{7})$$

$$=(\dfrac{5}{7})(\dfrac{4}{7})(\dfrac{4}{7})+(\dfrac{5}{7})(\dfrac{3}{7})(\dfrac{3}{7})+(\dfrac{2}{7})(\dfrac{4}{7})(\dfrac{3}{7})+(\dfrac{5}{7})(\dfrac{4}{7})(\dfrac{3}{7})$$

$$=\dfrac{(5\times 4\times 4)+(5\times 3\times 3)+(2\times 4\times 3)+(5\times 4\times 3)}{7 \times 7\times 7 }$$

$$=\dfrac{209}{243}$$

Answer is option (A)
Question 9
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A student appears for tests I, II and III. The student is considered successful if he passes in tests I, II or I, III or all the three. The probabilities of the student passing in tests I, II and III are m, n and $$\dfrac{1}{2}$$ respectively. If the probability of the student to be successful is $$\dfrac{1}{2}$$, then which one of the following is correct?

A
$$m(1 + n) = 1$$

B
$$n (1 + m) = 1$$

C
$$m = 1$$

D
$$mn = 1$$

Solution

Let $$A,B$$ and $$C$$ be the event that the student passes in tests I,II and III respectively.
$$P(A)=m, P(B)=n,P(C)=\dfrac{1}{2}$$

Probability that the student is successful $$=P(A\cap B \cap \overline{C}) +P(A\cap \overline{B}\cap C )+P(A\cap B\cap C)$$

$$\therefore \dfrac{1}{2}=P(A)P(B)P(\overline{C}) +P(A)P(\overline{B})P(C)+P(A)P(B)P(C)$$ ...($$A,B$$ and $$C$$ are independent events)

$$\therefore \dfrac{1}{2}=(m)(n)\left(1-\dfrac{1}{2}\right)+(m)(1-n)\left(\dfrac{1}{2}\right)+(m)(n)\left(\dfrac{1}{2}\right)$$

$$\therefore \dfrac{1}{2}=mn+\dfrac{m(1-n)}{2}$$

$$\therefore 1=mn+m$$

$$\therefore 1=m(1+n)$$

Answer is option (A)
Question 10
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An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. The probability that the person is a scooter driver is

A
$$\dfrac{1}{52}$$

B
$$\dfrac{3}{52}$$

C
$$\dfrac{15}{52}$$

D
$$\dfrac{19}{52}$$

Solution

Solution:
Let $$P(A)=P(scooter)=\cfrac{2000}{12000}=\cfrac16$$
$$P(B)=P(car)=\cfrac{4000}{12000}=\cfrac13$$
and $$P(C)=P(truck)=\cfrac{6000}{12000}=\cfrac12$$
Let $$E=$$ Event the person meets with an accident.
Then, $$P(E/A)=\cfrac{1}{100}, P(E/B)=\cfrac{3}{100},P(E/C)=\cfrac{15}{100}$$
Now, $$P(A/E)=\cfrac{P(A).P(E/A)}{P(A).P(E/A)+P(B).P(E/B)+P(C).P(E/C)}$$
$$=\cfrac{\cfrac{1}{6}\times\cfrac{1}{100}}{\cfrac16\times\cfrac1{100}+\cfrac13\times\cfrac3{100}+\cfrac12\times\cfrac{15}{100}}$$
$$=\cfrac{\cfrac16}{\cfrac16+1+\cfrac{15}2}$$
$$=\cfrac1{52}$$
Hence, A is the correct option.

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Re-Attempt Weekly Quiz Competition

Probability Test - 32

Result

Probability Test - 32

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SHARING IS CARING

Question 1

$$\dfrac{1}{1024}$$

$$\dfrac{243}{1024}$$

$$\dfrac{1023}{1024}$$

$$\dfrac{781}{1024}$$

Solution

Question 2

$$0.06$$

$$0.16$$

$$0.84$$

$$0.94$$

Solution

Question 3

$$1 - \left( \dfrac{1}{4} \right )^7$$

$$\left( \dfrac{1}{4} \right )^7$$

$$\left( \dfrac{3}{4} \right )^7$$

$$1- \left( \dfrac{3}{4} \right )^7$$

Solution

Question 4

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

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

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

$$\dfrac{1}{24}$$

Solution

Question 5

$$\dfrac{17}{18}$$

$$\dfrac{11}{12}$$

$$\dfrac{1}{12}$$

$$\dfrac{1}{18}$$

Solution

Question 6

$$0.98$$

$$0.91$$

$$0.70$$

$$0.49$$

Solution

Question 7

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

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

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

$$\dfrac{4}{7}$$

Solution

Question 8

$$\dfrac{209}{343}$$

$$\dfrac{134}{343}$$

$$\dfrac{149}{343}$$

$$\dfrac{60}{343}$$

Solution

Question 9

$$m(1 + n) = 1$$

$$n (1 + m) = 1$$

$$m = 1$$

$$mn = 1$$

Solution

Question 10

$$\dfrac{1}{52}$$

$$\dfrac{3}{52}$$

$$\dfrac{15}{52}$$

$$\dfrac{19}{52}$$

Solution

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