3 Marks Question

Question 513 Marks

Let $A$ and $B$ be two independent events such that $P(A) = p_1$ and $P(B) = p_2.$ Describe in words the events whose probabilities are: $1 - (1 - p_1)(1 - p_2).$

Answer

As, $1 − (1 − p_1)(1 − p_2) = 1 - [1 - P(A)] \times [1 - P(B)]$
$=1-\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})$
And, $A$ and $B$ are independent events.
i.e., $\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$\Rightarrow\ 1-(1-\text{p}_1)(1-\text{p}_2)=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\text{A}\cup\text{B})$
So, $\text{P}(\text{A}\cup\text{B})=1-(1-\text{p}_1)(1-\text{p}_2)$
Hence, $1 - (1 - p_1) (1 - p_2) = P ($At least one of $A$ and $B$ occurs$).$

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Question 523 Marks

Of the students in a college, it is known that $60\%$ reside in a hostel and $40\%$ do not reside in hostel. Previous year results report that $30\%$ of students residing in hostel attain $A$ grade and $20\%$ of ones not residing in hostel attain $A$ grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an $A$ grade. What is the probability that the selected student is a hosteler?

Answer

Let $A, E_1$ and $E_2$ denote the events that the selected student attains grade $A,$ resedes in a hostel and does not reside in a hostel, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{60}{100}$
$\text{P}(\text{E}_1)=\frac{40}{100}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{30}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{20}{100}$
Using Baye's theorem,
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{60}{100}\times\frac{30}{100}}{\frac{60}{100}\times\frac{30}{100}+\frac{40}{100}\times\frac{20}{100}}$
$=\frac{18}{18+8}=\frac{9}{13}$

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Question 533 Marks

Assume that the chances of a patient having heart attack is $40\%.$ It is also assumed that meditation and yoga course reduces the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%.$ At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer

Let $A, E_1,$ and $E_2$ respectively denote the events that a person has heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
$\therefore\ \text{P(A)}=0.40$
$\text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1)=04.\times0.70=0.28$
$\text{P}(\text{A}|\text{E}_2)=0.40\times0.75=0.30$
Probability that the patient suffering a heart attack followe a course of meditation and yoga is given by $P(E_1|A).$
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+\frac{1}{2}\times0.30}$
$=\frac{14}{29}$

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Question 543 Marks

From a pack of 52 cards, two are drawn one by one without replacement. Find the probability that both of them are kings.

Answer

A = first card is king
B = Second card is also king
Probability of getting two kings (Without replacement)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$ [Since, 4 kings out of 52 cards.]
$=\frac{1}{13}\times\frac{1}{17}$
$=\frac{1}{221}$
Required probability $=\frac{1}{221}$

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Question 553 Marks

In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that a student knows the answer given that he answered it correctly?

Answer

Let A, $E_1$ and $E_2$ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{3}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{4}$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{4}\times1}{\frac{3}{4}\times1+\frac{1}{4}\times\frac{1}{4}}$
$=\frac{3}{3+\frac{1}{4}}=\frac{12}{13}$

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Question 563 Marks

A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.

Answer

A bag contains 8 red and 6 green balls.
Three balls are drawn without replacement
P (At least 2 balls are green)
$=\text{P}\big[(\text{G}_1\cap\text{G}_2\cap\text{R}_1)\cup(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\\cup(\text{R}_1\cap\text{G}_1\cap\text{G}_2)\cup(\text{G}_1\cap\text{G}_2\cap\text{G}_3)\big]$
$=\text{P}\big(\text{G}_1\cap\text{G}_2\cap\text{R}_1)+\text{ P}(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\+\text{ P}(\text{R}_1\cap\text{G}_1\cap\text{G}_2)+\text{ P}(\text{G}_1\cap\text{G}_2\cap\text{G}_3)$
$=\text{P}\big(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{P}\Big(\frac{\text{R}_1}{\text{G}_1\cap\text{G}_2}\Big)\\+\text{P}(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{ P}\Big(\frac{\text{G}_3}{\text{G}_1\cap\text{G}_2}\Big)$
$=\frac{6}{14}\times\frac{5}{13}\times\frac{8}{12}+\frac{6}{14}\times\frac{8}{13}\times\frac{5}{12}\\+\frac{8}{14}\times\frac{6}{13}\times\frac{5}{12}+\frac{6}{14}\times\frac{5}{13}\times\frac{4}{12}$
$=\frac{1}{14}\times\frac{1}{13}\times\frac{1}{12}\times(240+240+240+120)$
$=\frac{840}{14\times13\times12}$
$=\frac{5}{13}$
Required probability $=\frac{5}{13}$

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Question 573 Marks

Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Answer

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Consider the given events.
A = An even number on the card
B = A number more than 3 on the card
Clearly,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}
Now,
$\text{A}\cap\text{B}=\{4, 6, 8, 10\}$
$\therefore \text{Required probability} =\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{4}{7}$

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Question 583 Marks

If P(A) = 0.4, P(B) = 0.3 and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5$ find $\text{P}(\text{A}\cap\text{B})$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$

Answer

Given,
P(A) = 0.4, P(B) = 0.3 and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5$
We know that,
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.4}$
$\text{P}(\text{A}\cap\text{B})=0.2$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.2}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{3}$

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Question 593 Marks

A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Answer

Here two dice are thrown
A = Getting 7 as sum on two dice
A {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = Second die exhibits an odd number
B = {(1,1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (5, 5)}
$(\text{A}\cap\text{B})=\{(2, 5), (4, 3), (6, 1)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{18}$
$=\frac{1}{6}$
Hence, Required probability $=\frac{1}{6}$

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Question 603 Marks

X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in two subject.

Answer

Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A grade in two subjects)
$=\text{P}\big((\text{m}\cap\text{p}\cap\overline{\text{c}})\cup(\text{m}\cap\overline{\text{p}}\cap\text{c})\cup(\overline{\text{m}}\cap\text{p}\cap\text{c})\big)$
$=\text{P}(\text{m})\text{P}(\text{p})\text{P}(\overline{\text{c}})+\text{P}(\text{m})\text{P}(\overline{\text{p}})\text{P}(\text{c})+\text{P}(\overline{\text{m}})\text{P}(\text{p})\text{P}(\text{c})$
$=\text{P}(\text{m})\text{P}(\text{p})(1-\text{P(c)})+\text{P(m)}(1-\text{P(p)})\text{p(c)}+(1-\text{p(m)})\text{P(p)}\text{P(c)}$
$=(0.2)(0.3)(1-0.5)+(0.2)(1-0.3)(0.5)+(1-0.2)(0.3)(0.5)$
$=(0.2)(0.3)(0.5) + (0.2)(0.7)(0.5)+(0.8)(0.3)(0.5)$
$=0.03+0.07+0.12$
$=0.22$
Required probability = 0.22

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Question 613 Marks

An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?

Answer

Urn contains 3 white, 4 red and 5 black balls. Total balls = 12
Two balls are drawn without replacement
A = First ball is black
B = Second ball is black
P (Atleast one ball is black)
$=\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P }\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)$
$=1-\Big(\frac{7}{12}\times\frac{6}{12}\Big)$
$=1-\frac{7}{22}$
Required probability $=\frac{15}{22}$

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MATHS 3 Marks Question