Probability

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Questions:

Generic 3

12 x 3 = 36

In an examination, an examinee either guesses or copies or knows the answer o multiple choice questions with four choices. The probability that he makes a guess is \(1\over3\) and probability that he copies the answer is \(1\over6\). The probability that his answer is correct, given that he copied it, is \(1\over8\). Find the probability that he knew the answer to the question, given that he correctly answered it.

The probability of a student A passing an examination is \(3\over5\) and of student B is \(4\over5\).Assuming that the two events "A passes", "B passes" as independent. Find the probability of:
(i) Both the students passing the examination
(ii) Only A passing the examination
(iii) Only of them passing the examination
(iv) none of them passing the examination.

In a group of 50 scouts in a camp, 30 are well trained in first aid techniques while the remaining are well trained in hospitality but not in first aid. Two scouts are selected at random from the group, Find the probability distribution of number of selected scouts who are well trained in first aid. Find the mean of the distribution also. Write one more value which is expected from a well trained scout.

In a certain college 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75metres. Find the probability that the selected students is girl?

How many times must a man toss a fair coin, so that the probability of having at least one head is more than 90%?

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all the balls in the bag are white?

A pair of dice is thrown 4 times. If getting a doublet is considered a success find the mean and variance of the number of successes.

A and B throw a pair of die turn by turn. The first to throw 9 is awarded a prize. If A starts the game, show that the probability of A getting the prize is \(9\over17 \)

10.

12 cards, numbered 1 to 12, are placed in box mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, find the probability that it is an even number.

11.

There are 2000 scooter drivers, 4000 car drivers and 6000 truck drivers all insured. The probabilities of an accident involving a scooter, a car, a truck are 0.01, 0.03, 0.15 respectively. One of the insured drivers meets with an accident. What is the probability that he is a scooter driver?

12.

Find the binomial distribution for which mean is 4 and variance 3.

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Answers Key:

Generic 3

12 x 3 = 36

A: guesses answer \(P(A)=\frac{1}{3}\)
B: copies answer \(P(B)=\frac{1}{6}\)
C: knows answer \(P(C)=1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}\)
E: answer correctly
\(P(E / A)=\frac{1}{4} \text { (multiple choice question with four choices); }\)
\(P(E / B)=\frac{1}{8} \text { and } P(E / C)=1\)
Using Bayes' Theorem probability that he knows answer to the question given that he correctly answered it is
\(P(C / E)=\frac{P(C) \cdot P(E / C)}{P(A) \cdot P(E \mid A)+P(B) \cdot P(E / B)+P(C) \cdot P(E / C)}\)
\(=\frac{\frac{1}{2} \times 1}{\frac{1}{3} \times \frac{1}{4}+\frac{1}{6} \times \frac{1}{8}+\frac{1}{2} \times 1}=\frac{24}{4+1+24}=\frac{24}{29}\)

(i) P(both the students passing the examination)
\(=\frac{3}{5} \times \frac{4}{5}=\frac{12}{25}\)
(ii) P( only student A passing the examination)
\(=\frac{3}{5} \times \frac{1}{5}=\frac{3}{25}\)
(iii) P( only one of them passing the examination)
= p(A passes and B does not pass) or (A does not pass and B passes)
\(=\frac{3}{5} \times \frac{1}{5}+\frac{2}{5} \times \frac{4}{5}=\frac{3+8}{25}=\frac{11}{25}\)
(iv) P(none of them passing the examination)
= P (A does not pass and B does not pass)
\(=\frac{2}{5} \times \frac{1}{5}=\frac{2}{25}\)

1.2 Another value expected from a well trained scout is brave/pure in thoughts and deeds.

\(3\over11\)

Let the coin be lossed n times.
P(getting a head) = \(\frac { 1 }{ 2 } \)( = p)
P(not getting no head) = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)(=q)
P(at lest one head) = 1-P(0)
= \(1-^{ n }{ C }_{ 0 }{ q }^{ n }{ p }^{ 0 }=1-\left( 1 \right) { \left( \frac { 1 }{ 2 } \right) }^{ n }\left( 1 \right) \)
= \(1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
By the question, \(1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }>\frac { 90 }{ 100 } \)
\(\Rightarrow 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }>0.9\Rightarrow { \left( \frac { 1 }{ 2 } \right) }^{ n }<1-0.9\)
\(\Rightarrow 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n }<0.1\Rightarrow n\ge 4\)

Let the events be:
E₁ : Item from machine A
E₂ : Item from machine B
And A : Item is defective.
\(P({ E }_{ 1 })=\frac { 60 }{ 100 } =\frac { 3 }{ 5 } \)
\(P({ E }_{ 2 })=\frac { 40 }{ 100 } =\frac { 2 }{ 5 } \)
Also \(P(A/{ E }_{ 1 })=\frac { 2 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 1 }{ 100 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) }{ \left( \frac { 3 }{ 5 } \right) \left( \frac { 2 }{ 100 } \right) +\left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) } \)
\(=\frac { 2 }{ 6+2 } =\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)

Let A :Two drawn balls are white
E₁ : All the balls are white
E₂ : Three balls are white
E₃ : Two balls are white
Since, E₁, E₂, and E₃ are mutually exclusive and exhaustive events.
\(\therefore \quad P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}\)
Now, \(P\left(\frac{A}{E_1}\right)=\frac{{ }^4 C_2}{{ }^4 C_2}=1, P\left(\frac{A}{E_2}\right)=\frac{{ }^3 C_2}{{ }^4 C_2}=\frac{3}{6}=\frac{1}{2}\)
and \(P\left(\frac{A}{E_3}\right)=\frac{{ }^2 C_2}{{ }^4 C_2}=\frac{1}{6}\)
\(\therefore\) Probability that all balls in the bag are white
\(P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{\left[P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)\right.}\)\(\left.+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)\right]\)
\(=\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1+\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times \frac{1}{6}}\)\(=\frac{1}{1+\frac{1}{2}+\frac{1}{6}}=\frac{6}{10}=0.6\)

mean = 2/3
variance = 5/9

S: getting a total of = {(3, 6), (4, 5), (5, 4), (6, 3)}
\(P(S)=\frac{4}{36}=\frac{1}{9} \cdot P(\bar{S})=\frac{8}{9}\)
A can win in 1^st, 3^rd, 5^th, 7^th, ..... throws
\(P(A) =P(S)+[P(\bar{S})]^{2} P(S)+[P(\bar{S})]^{4} P(S)+\cdots \cdot \)
\(=\frac{1}{9}+\left(\frac{8}{9}\right)^{2} \cdot \frac{1}{9}+\left(\frac{8}{9}\right)^{4} \cdot \frac{1}{9}+\ldots \ldots \)
\(=\frac{\frac{1}{9}}{1-\frac{64}{81}}=\frac{9}{17} \quad\left[\begin{array}{l} \text { sum of infinite GP } \\ a+a r+a r^{2}+\ldots=\frac{a}{1-r} \end{array}\right]\)

10.

Total cards are 12
A : number drawn is more than 3, i.e. 4, 5, 6, ..., 12.
B : getting an even number, i.e. 2, 4, 6, 8, 10, 12.
\(A \cap B: 4,6,8,10,12 \)
\(P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{5 / 12}{9 / 12}=\frac{5}{9} . \)

11.

E : accident; S : scooter driver; C : car driver; T: truck driver
\(P(S)=\frac{2000}{12000}=\frac{2}{12} ; P(C)=\frac{4000}{12000}=\frac{4}{12} ; \)
\(P(T)=\frac{6000}{12000}=\frac{6}{12}\)
P(EIS) = 0.01; P(EIC) = 0.03; P(EIT) = 0.15
Using Bayes' Theorem the probability of accident of a scooter driver is
\(P(S / E) =\frac{P(S) \cdot P(E / S)}{P(S) \cdot P(E / S)+P(C) \cdot P(E / C)+P(T) \cdot P(E / T)} \)
\(=\frac{\frac{2}{12} \times 0.01}{\frac{2}{12} \times 0.01+\frac{4}{12} \times 0.03+\frac{6}{12} \times 0.15} \)
\(=\frac{0.02}{0.02+0.12+0.90}=\frac{2}{104}=\frac{1}{52} \)

12.

Given, mean = np = 4, Also, variance = npq = 3
\(\Rightarrow q=\frac{3}{4} \text { and } p=1-q=1-\frac{3}{4}=\frac{1}{4}, n=16\)
\(\therefore \text { distribution is }(q+p)^{n}=\left(\frac{3}{4}+\frac{1}{4}\right)^{16}\)