Solve the Following Question.(4 Marks)

Question 14 Marks

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that

Both the balls are red.
One ball is red and the other is black.
One ball is white.

Answer

Urn7-White balls
5-Black balls
3-Red balls
Since two balls are drawn at random
$\therefore\text{n}(\text{S})=\frac{15}{2}$

E be the event that both the balls are red

$\therefore\text{n}(\text{S})=^3\text{C}_2$
$\therefore\text{p}(\text{E})=\frac{^3\text{C}_2}{^{15}\text{C}_2}=\frac{3\times2}{15\times14}=\frac{1}{35}$

E be the event that one ball is red and other is black

$\therefore\text{n}(\text{E})=^3\text{C}_1\times^5\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^3\text{C}_1\times^5\text{C}_1}{^{15}\text{C}_2}$
$=\frac{3\times5\times2}{15\times16}=\frac{1}{7}$

E be the event that one ball is white

$\therefore\text{n}(\text{E})=^7\text{C}_1\times^8\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^7\text{C}_1\times^8\text{C}_1}{^{15}\text{C}_2}$
$=\frac{7\times6\times2}{14\times15}=\frac{8}{15}$

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Question 24 Marks

In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?

Answer

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=0.87+0.36-0.30=0.93$

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Question 34 Marks

Five cards are drawn from form a pack of 52 cards. what is the chance that these 5 will contain:

Just one ace
At least one ace.

Answer

Since five card are drawn from a pack to 52 card
$=^{52}\text{C}_5$

Let E be the event that those five card contain exactly one ace.

$\therefore\text{n}(\text{E})=\ ^{4}\text{C}_1\times\ ^{48}\text{C}_4$

$\therefore\text{n}(\text{E})=\frac{^4\text{C}_1\times^{48}\text{C}_4}{^{52}\text{C}_5}$

$=\frac{4\times48\times47\times46\times45}{\frac{52\times51\times50\times49\times48}{5}}$

$=\frac{3243}{10829}$

Let E be the event that five cards contain atleast one ace.

$\therefore\text{E}= \big\{{1\ \text{or}\ 2\text{ or } 3 \text{ or }4}\big\}$

$\text{n}(\text{E})=\frac{^4\text{C}_1\times^{48}\text{C}_4\ +\ ^4\text{C}_2\times^{48}\text{C}_3\ +\ ^4\text{C}_3\times^{48}\text{C}_2\ +\ ^4\text{C}_4\times^{48}\text{C}_1}{^{52}\text{C}_5}$

$=\frac{4\times\frac{48\times47\times46\times45}{4\times3\times2\times1}+\frac{4\times3}{2}\times\frac{48\times47\times46}{3\times2\times1}+4\times\frac{48\times47}{2}+48}{\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}}$

$=\frac{18472}{54145}$

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Question 44 Marks

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English Examination is 0.75. What is the probability of passing the Hindi Examination?

Answer

Let E be the event that student passed in english examination
$\therefore\text{P}(\text{E})=0.75$
Let H be the event that student passed in hindi examination
$\therefore\text{P}(\text{H})=?$
Also, $\text{P}(\text{E}\cap\text{H})=0.5$ and $\text{P}(\overline{\text{E}\cap\text{H}})=0.1$
$\because\text{P}(\overline{\text{E}}\cap\overline{\text{H}})=1-\text{P}(\text{E}\cup\text{H})$
$\Rightarrow\text{P}({\text{E}}\cup{\text{H}})=1-0.1$
$=0.9$
Now,
$\text{P}({\text{E}}\cup{\text{H}})=\text{P}(\text{E})+\text{P}(\text{H})-\text{P}({\text{E}}\cap{\text{H}})$
$0.9=0.75+\text{P}(\text{H})-0.5$
$\text{P}(\text{H})=0.90-0.25$
$=0.65$

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Question 54 Marks

Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that either both are black or both are kings.

Answer

Two cards are drawn from a well shuffled deck of cards
$\therefore\text{n}(\text{S})=^{52}\text{C}_2$
A be the event of getting black cards
$\text{n}(\text{A})=^{26}\text{C}_2$ [$\because$ There are 26 black cards]
$\therefore\text{p}(\text{A})=\frac{^{26}\text{C}_2}{^{52}\text{C}_2}$
$=\frac{26\times25}{52\times51}$
B be to event of getting both king cards
$\therefore\text{p}(\text{B})=\frac{^{4}\text{C}_2}{^{52}\text{C}_2}$
$=\frac{4\times3}{52\times51}$ [$\because$ There are 4 king cards]
Also, $\text{p}(\text{A}\cap\text{B})=\frac{^2\text{C}_2}{^{52}\text{C}_2}$
$=\frac{2\times1}{52\times51}$ [$\because$ Two king are black also]
Now,
$\text{p}(\text{A}\cup\text{B})=\text{p}(\text{A})+\text{p}(\text{B})-\text{p}(\text{A}\cap\text{B})$
$=\frac{26\times25}{52\times51}+\frac{4\times3}{52\times51}-\frac{2}{52\times51}$
$=\frac{660}{52\times51}$
$=\frac{55}{221}$

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Question 64 Marks

A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Answer

Since a number is chosen from first 500
$\therefore\text{n}(\text{S})=500$
Let A be the event of choosing number divisible by 3 = [3, 6, 9, ...., 498]
$\therefore\text{n}(\text{A})=166$
$\text{P}(\text{A})=\frac{166}{500}$ $\begin{bmatrix}\because\text{a}+(\text{n}-1)\text{d}=498\\3+(\text{n}-1)3=498\\3\text{n}=498\Rightarrow\text{n}=166\end{bmatrix}$
B the event of choosing number
[5, 10, 15, ..., 500]
$\text{n}(\text{B})=100$
$\Rightarrow\text{P}(\text{B})=\frac{100}{500}$
Also, $\text{P}({\text{A}\cap\text{B}})=\{15,\ 30,\ ...,\ 495\}$
$\Rightarrow\text{n}({\text{A}\cap\text{B}})=33$
$\text{P}({\text{A}\cap\text{B}})=\frac{33}{500}$
$\therefore\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{166}{500}+\frac{100}{500}-\frac{33}{500}$
$=\frac{233}{500}$

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Question 74 Marks

If A and B are mutually exclusive events such that P(A) = 0.35 and P(B) = 0.45, find:

$\text{P}(\text{A}\cup\text{B})$
$\text{P}({\text{A}}\cap{\text{B}})$
$\text{P}({\text{A}}\cap\overline{\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

Answer

It is given that A and B are mutually exclusive events.
$\therefore\text{P}(\text{A}\cap\text{B})=0$
Also, P(A) = 0.35 and P(B) = 0.45.

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$=0.35+0.45-0$

$=0.80$

A and B are mutually exclusive events.

$\therefore\text{P}({\text{A}}\cap{\text{B}})=0$

$\text{P}({\text{A}}\cap\overline{\text{B}})=\text{p}(\text{A})-\text{P}({\text{A}}\cap{\text{B}})$

$=0.35-0$

$=0.35$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup{\text{B}}})$

$=1-\text{P}({\text{A}\cup{\text{B}}})$

$=1-0.80$

$=0.20$

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Question 84 Marks

Find the probability of getting 2 or 3 tails when a coin is tossed four times.

Answer

$\because$ A coin is tossed four times
$\Rightarrow\text{n}(\text{S})=2^4=16$
Let A be the event of getting 2 tails
$\therefore\text{A}=\{\text{HHTT},\ \text{HTHT},\ \text{HTTH},\ \text{THTH},\ \text{TTHH},\ \text{THHT}\}$
$\therefore\text{p}(\text{A})=\frac{6}{16}$
Let B be the event of getting 3 tails,
$\therefore\text{B}=\{\text{HTTT},\ \text{THTT},\ \text{TTHT},\ \text{TTTH\}}$
$\therefore\text{p}(\text{B})=\frac{4}{16}$
$\because$ A and B are mutually exclusive case.
$\therefore\text{n}(\text{A}\cup\text{B})=\text{n}(\text{A})+\text{n}(\text{B})$
$=\frac{6}{16}+\frac{4}{16}$
$=\frac{10}{16}$
$=\frac{5}{8}$

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Question 94 Marks

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Answer

Number of multiples of 2 in 1 to 1000 are 500
Number of multiples of 9 in 1 to 1000 are 111
Out of 111, 55 are even numbers. So total favorables number are
500 + 56 = 556
Probability that integer is a multiple of 2 or a multiple of 9
$=\frac{556}{1000}=0.556$

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Question 104 Marks

An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.

Answer

Let A be the event of choosing a positive integer divisible by 6
$\therefore\text{A}=\{6,\ 12,\ ....,\ 198\}$
$\Rightarrow\text{n}(\text{A})=33$
$\therefore\text{p}(\text{A})=\frac{33}{200}$
Let B be the event of choosing a positive integer divisible by 8
$\therefore\text{B}=\{8,\ 16,\ ....,\ 200\}$
$\Rightarrow\text{n}(\text{B})=25$
$\therefore\text{p}(\text{B})=\frac{25}{200}$
Also, $\text{A}\cap\text{B}=\{24,\ 28,\ ...,\ 192\}$
$\Rightarrow\text{n}(\text{A}\cap\text{B})=8$
$\therefore\text{p}(\text{A}\cap\text{B})=\frac{8}{200}$
$\therefore\text{p}(\text{A}\cup\text{B})=\frac{1}{4}$

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Maths Solve the Following Question.(4 Marks)

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