(Each question 4 marks) - MATHS STD 11 Science Questions - Vidyadip

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15 questions · self-marked practice — reveal the answer and mark yourself.

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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Question 114 Marks

A bag contains 6 red, 4 white and 8 blue balls. if three balls are drewn at random, find the probability that:

(i) One is red and two are white

(ii) Two are blue and one is red

(iii) One is red.

Answer

BAG: 6 - Red ball, 4 - White ball, 8 - blue ball Since three ball are drawn, $\therefore\text{n}\text{(S)}=\ ^{18}\text{C}_3$ (i) Let E be the events that one red and two white ball are drawn. $\therefore\text{n}\text{(S)}=\ ^{6}\text{C}_1\times\ ^{4}\text{C}_2$ $\therefore\text{P}\text{(E)}=\frac{\ ^{6}\text{C}_1\times\ ^{4}\text{C}_2}{\ ^{18}\text{C}_3}=\frac{6\times4\times3}2\times{\frac{3\times2}{18\times17\times16}}$ $\text{P}\text{(E)}=\frac{3}{68}$ (ii) Let E be the event that two blue and one red ball are drawn. $\therefore\text{n}\text{(E)}=\ ^{8}\text{C}_2\times\ ^{6}\text{C}_1$ $\therefore\text{P}\text{(E)}=\frac{\ ^{8}\text{C}_2\times\ ^{4}\text{C}_2}{\ ^{18}\text{C}_3}=\frac{8\times7}2\times6\times{\frac{3\times2\times1}{18\times17\times16}}=\frac{7}{34}$ $\text{P}\text{(E)}=\frac{7}{34}$ (iii) Let E be the event thatone of the ball must be red. $\therefore\text{E}=\big\{(\text{R, W, B)}\ \text{or}\ \big\{(\text{R, W, W)}\ \text{or}\ \big\{(\text{R, B, B)}$ $\therefore\text{n}\text{(E)}=\ ^6\text{C}_1\times\ ^4\text{C}_1\times \ ^8\text{C}_1+\ ^6\text{C}_1\times \ ^4\text{C}_2+\ ^6\text{C}_1\ \times\ ^8\text{C}_2$ $\therefore\text{P}\text{(E)}=\frac{\ ^6\text{C}_1\times\ ^4\text{C}_1\times \ ^8\text{C}_1+\ ^6\text{C}_1\times \ ^4\text{C}_2+\ ^6\text{C}_1\ \times\ ^8\text{C}_2}{\ ^{18}\text{C}_3}$ $=\frac{396}{816}=\frac{33}{68}$

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Question 124 Marks

A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. find the probability that:

All 10 are defective
All 10 are good
At least one is defective
None is defective

Answer

A box contains 100 bulbs, out of which 20 are defective, $\therefore$ Numbers of good bulbs 100 - 20 = 80 Now, 10 balls are selected for inspection $\therefore$ Numbers of elementary events in sample space $\text{n}\text{(S)}=\ ^{100}\text{C}_{10}$

Let E be the events that all 10 bulbs selected are defective

$\text{n}\text{(E)}=\ ^{20}\text{C}_{10}$
$\therefore\text{P}\text{(E)}=\frac{\ ^{20}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$
$=\frac{\ ^{20}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events that all 10 bulbs good bulbs are selected

$\therefore\text{n}\text{(E)}=\ ^{80}\text{C}_{10}$
$\therefore\text{P}\text{(E)}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events thatatleast one bulbs is defective

$\text{E}=\big\{1,\ 2,\ 3,\ 4,\ 5, \ 6,\ 7,\ 8,\ 9,\ 10\big\}$
Where,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, are numbers of defective bulbs
$\because\ \stackrel{{\sim}}{\hbox{E}}$ be the event that none of the bulbs are defective
$\therefore \text{n}\stackrel{{\sim}}{\hbox{(E)}}=\ ^{80}\text{C}_{10}$
$\therefore \text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$
$\therefore\text{P}\text{(E)}=1-\text{P}\stackrel{{\sim}}{\hbox{(E)}}$
$=1-\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events thatatleast one bulbs is defective, that is all bulbs are good so,

$\text{n}\text{(E)}=\ ^{80}\text{C}_{10}$
$\text{P}\text{(E)}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

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Question 134 Marks

20 cards are numbered form 1 to 20. card is drawn at random. what is the probability that trhe number on the card is:

A multiple of 4?
Not a multiple of 4?
odd?
Greather than 12?
Divisible by 5?
Not a multiple of 6?

Answer

We have 20 cards numbered from 1 to 20, one card is drawn at random $\therefore\text{n(S)}=\ ^{20}\text{C}_1=20$

Let E be the event that the number on the drawn cards is multiple of 4

$\therefore\text{E}=\big\{4, \ 8, \ 12,\ 16,\ 20\big\}$
$\therefore\text{n(E)}=5$
$\therefore\text{p(E)}=\frac{5}{20}=\frac{1}{4}$

Let E be the event that the number on the drawn cards is not multiple of 4

$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that the number on the drawn cards is not multiple of 4
$\therefore\ \stackrel{{\sim}}{\hbox{E}}\ =\big\{4,\ 8, \ 12, \ 16,\ 20\big\}$
$\Rightarrow\text{n}\stackrel{{\sim}}{\hbox{(E)}}=5$
$\therefore\text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{5}{20}=\frac{1}{4}$
$\therefore\text{P(E)}=1-\text{P}(\stackrel{{\sim}}{\hbox{E)}}$
$=1-\frac{1}{4}=\frac{3}{4}$

Let E be the event that the number on the drawn cards is odd.

$\therefore\text{E}=\big\{1, \ 3,\ 5,\ 7,\ 13,\ 15,\ 17,\ 19\big\}$
$\therefore\text{n(E)}=10$
$\Rightarrow\text{P(E)}=\frac{10}{20}=\frac{1}{2}$

Let E be the event that the number on the drawn cards is greater than 12.

$\therefore\text{E}=\big\{13, \ 14,\ 15,\ 16,\ 17,\ 18,\ 19,\ 20\big\}$
$\therefore\text{n(E)}=8$
$\Rightarrow\text{P(E)}=\frac{8}{20}=\frac{2}{5}$

Let E be the event that the number on the drawn cards is divisible by 5.

$\therefore\text{E}=\big\{5, \ 10,\ 15,\ 20\big\}$
$\text{n(E)}=4$
$\therefore\text{P(E)}=\frac{4}{20}=\frac{1}{5}$

Let E be the event that the number on the drawn cards is divisible by 6.

$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that the number on the drawn cards is not divisible of 6
$\Rightarrow\text{n}\stackrel{{\sim}}{\hbox{(E)}}=3$
$\therefore\text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{3}{20}$
$\text{P(E)}=1-\text{P}(\stackrel{{\sim}}{\hbox{E)}}$
$=1-\frac{3}{20}=\frac{17}{20}$

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Question 144 Marks

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. from the box, what is the probability that:

All are blue?
At least one is green?

Answer

10 Red, 20 Blue, 30 Green

All 5 are blue

$=\frac{\ ^{20}\text{C}_5\times\ ^{40}\text{C}_0}{\ ^{60}\text{C}_5}$ $=\frac{34}{11977}$

Atleast one green = 1 - no green

Different combination one possible for no green case are 5B, 1R, 2R, 3B, 3R, 2B, 4R, 2B, 5R,
$5\text{B}=\ ^{20}\text{C}_5$
$1\text{R}\ 4\text{B}=\ ^{10}\text{C}_1\times\ ^{20}\text{C}_4$
$2\text{R}\ 3\text{B}=\ ^{10}\text{C}_2\times\ ^{20}\text{C}_3$
$3\text{R}\ 2\text{B}=\ ^{10}\text{C}_3\times\ ^{20}\text{C}_2$
$4\text{R}\ 1\text{B}=\ ^{10}\text{C}_1\times\ ^{20}\text{C}_1$
$5\text{R}=\ ^{10}\text{C}_5$
Atleast one green = 1 - no green
$=1-\frac{\ ^{20}\text{C}_5+\ ^{10}\text{C}_1\times\ ^{20}\text{C}_4+\ ^{10}\text{C}_2\times\ ^{20}\text{C}_3+\ ^{10}\text{C}_3\times\ ^{20}\text{C}_2+\ ^{10}\text{C}_4\times\ ^{20}\text{C}_1+\ ^{10}\text{C}_5}{\ ^{60}\text{C}_5}$
$=\frac{4367}{4484}$

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Question 154 Marks

A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has

All boys?
All girls?
1 boys and 2 girls?
At least one girl?
At most one girl?

Answer

10 boys 8 girls Three student are selected at random $\text{n}(\text{S})=^{18}\text{C}_3$

E be the event that the group has all boys

$\therefore\text{n}(\text{E})=^{18}\text{C}_3$
$\therefore\text{p}(\text{E})=\frac{^{10}\text{C}_3}{^{18}\text{C}_3}$
$=\frac{10\times9\times8}{18\times17\times16}$
$=\frac{5}{34}$

E be the event that the group has all girls

$\therefore\text{n}(\text{E})=^8\text{C}_3$
$\therefore\text{p}(\text{E})=\frac{^{8}\text{C}_3}{^{18}\text{C}_3}$
$=\frac{8\times7\times6}{18\times17\times16}$
$=\frac{7}{102}$

E be the event that the group has one boy and two girls

$\therefore\text{n}(\text{E})=^8\text{C}_3\times^{10}\text{C}_2$
$\therefore\text{p}(\text{E})=\frac{^{8}\text{C}_1\times^{10}\text{C}_2}{^{18}\text{C}_3}$
$=\frac{35}{102}$

E be the event that atleast one girls in the group

$\text{E}=(1,\ 2,\ 3)\text{girls}$
$\therefore\text{n}(\text{E})=^8\text{C}_1\times^{10}\text{C}_2+^8\text{C}_2\times^{10}\text{C}_1+^8\text{C}_3\times^{10}\text{C}_0$
$\text{p}(\text{E})=\frac{^{8}\text{C}_1\times^{10}\text{C}_2+^{8}\text{C}_2\times^{10}\text{C}_1+^{8}\text{C}_3}{^{18}\text{C}_3}$
$=\frac{29}{34}$

E be the event that almost one girls in the group

$\text{E}=(0,\ 1,)\text{girls}$
$\therefore\text{n}(\text{E})=^8\text{C}_0\times^{10}\text{C}_3+^8\text{C}_1\times^{10}\text{C}_2$
$\text{p}(\text{E})=\frac{^{10}\text{C}_3+8\times^{10}\text{C}_2}{^{18}\text{C}_3}$
$=\frac{10}{17}$

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