3 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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

Question 13 Marks

In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

Answer

Since two dice is thrown,
$\therefore\text{S}=6^2=36$
Let A be the event of choosing doublet
$=\big\{(1,\ 1),\ (2,\ 2),\ (3,\ 3),\ (4,\ 4),\ (5,\ 5),\ (6,\ 6)\big\}$
$\Rightarrow\text{P}(\text{A})=\frac{6}{36}=\frac{1}{6}$
B the event of choosing total of 9.
$=\big\{(3,\ 6),\ (4,\ 5),\ (5,\ 4),\ (6,\ 3)\big\}$
$\Rightarrow\text{P}(\text{B})=\frac{4}{36}=\frac{1}{9}$
$\therefore$ Probability of choosing neither a doublet nor a total of 9.
$=\text{P}(\overline{\text{A}\cap\text{B}})=1-\text{P}({\text{A}\cup\text{B}})\ ...(\text{i})$
Now,
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{6}+\frac{1}{9}+0$
$=\frac{3+2}{18}$
$=\frac{5}{18}$
Now,
$\text{P}(\text{A}\cup\text{B})=\frac{5}{18}$
$\therefore$ Simplies $\text{P}(\overline{\text{A}\cap\text{B}})=1-\frac{5}{8}$
$=\frac{13}{18}$

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Question 23 Marks

box contains 6 red marbles numbered 1 through 6 and 4 white marble numbered form 12 through 15. find the probability that a marble drawn is:

White
White and odd numbered
Even numbered
Red or even numbered.

Answer

We have 6 red marble numbered 1 - 6 and we have 4 white marble numbered 12 - 15 one marble is tobe drawn

$\therefore\text{n(s)}=\ ^{10}\text{C}_1$

E be event of getting white marble,

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

$\therefore\text{P(E)}=\frac{\ ^{4}\text{C}_1}{\ ^{10}\text{C}_1}=\frac{4}{10}=\frac{1}{5}$

E be the event of getting white marble with odd numbered marble.

$\therefore\text{E}=\big\{13,\ 15\big\}$

$\Rightarrow\text{n(E)}=2$

$\text{P(E)}=\frac{2}{10}=\frac{1}{5}$

E be the event of getting even numbered marble.

$\therefore\text{E}=\big\{2,\ 4,\ 6,\ 12,\ 24\big\}$

$\Rightarrow\text{n(E)}=5$

$\text{P(E)}=\frac{5}{10}=\frac{1}{2}$

E₁ be the event of getting red marble.

$\therefore\text{P}\text{(E)}_1=\frac{5}{10}$ [as in (ii)]

$\therefore\text{(E}_1\cap\text{E}_2)=\text{even numbered marble}=\big\{2,\ 4,\ 6\big\}$

$\Rightarrow\text{n}(\text{E}_1\cap\text{E}_2)=3$

$\Rightarrow\text{P}(\text{E}_1\cap\text{E}_2)=\frac{3}{10}$

$\therefore$ by law of addition,

$\Rightarrow\text{P}(\text{E}_1\cup\text{E}_2)=\text{P(E}_1)+\text{P(E}_2)-\text{P}(\text{E}_1\cap\text{E}_2)$

$=\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{8}{10}$

$=\frac{4}{5}$

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Question 33 Marks

In a lottary, a person chooses six different numbers at random from 1 to 20, and if these six numbers match with six numbers already fixed by the lottery commitee, he wins the prize what is a probability of winning the prize in the game?

Answer

As six number has been choose from 1 to 20 numbers

$\therefore\ ^{20}\text{C}_6$

Let E be the event that six number choose in matched with the given number

[As winning number is fixed]

$\Rightarrow\text{n}\text{(E)}=1$

$\therefore\text{P}\text{(E)}=\frac{1}{\ ^{20}\text{C}_6}$

$=\frac{6\times5\times4\times3\times2\times1}{20\times19\times18\times17\times16\times15}$

$=\frac{1}{38760}$

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Question 43 Marks

One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

Answer

Let A be the event of choosing a number divisible by 4

$\therefore\text{A}=\{4,\ 5,\ ...,\ 100\}$

$\Rightarrow\text{n}(\text{A})=25$

$\text{p}(\text{A})=\frac{25}{100}$

Let B be the event of choosing a number dividible by 6

$\therefore\text{B}=\{6,\ 12,\ ...,\ 96\}$

$\Rightarrow\text{n}(\text{B})=16$

$\text{p}(\text{B})=\frac{16}{100}$

Also, $(\text{A}\cap\text{B})=\{12,\ 24,\ ....,\ 96\}$

$\Rightarrow\text{n}(\text{A}\cap\text{B})=8$

$\therefore\text{P}({\text{A}}\cap{\text{B}})=\frac{8}{100}$

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

$=\frac{25}{100}+\frac{16}{100}-\frac{8}{100}$

$=\frac{33}{100}$

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Question 53 Marks

Form a deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honours of the same suit.

Answer

Since from Form a deck of cards, four cards are drawn
$\therefore\ \text{n}\text{(E)}=\ ^{52}\text{C}_4$
Let E be the event of that all four cards drawn are honour cards from same suit,
(Hounour cards means king, queen, jack & Ace)
$\therefore\text{E}=\ ^{4}\text{C}_4\ \text{or}\ \ ^{4}\text{C}_4\ \text{or}\ \ ^{4}\text{C}_4\ \text{or}\ ^{4}\text{C}_4\ $
$\Rightarrow\text{n}\text{(E)}=4\times\ ^4\text{C}_4=4$
$\therefore\text{P}\text{(E)}=\frac{4}{\ ^{52}\text{C}_4}$
$=\frac{4\times4\times3\times2\times1}{{52\times51\times50\times49}}$
$=\frac{96}{6497400}$
$=\frac{4}{270725}$

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Question 63 Marks

A big contains 6 red, 4 white and 8 blue balls, if three balls are drawn at random, find the probability that one is red, one is white and one is blue.

Answer

BAG: 6 - Red ball, 4 - White ball, 8 -Blue ball
$\because$ Three balls are drawn at random
$\therefore\text{n}\text{(S)}=\ ^{18}\text{C}_ 3$
Let E be the event that one red ball, one white ball and one blue ball was drawn.
$\therefore\ \text{n}\text{(E)}=\ ^{6}\text{C}_1\times\ ^{4}\text{C}_1\times\ ^{8}\text{C}_1$
$\therefore\ \text{P}\text{(E)}=\frac{\ ^{6}\text{C}_1\times\ ^{4}\text{C}_1\times\ ^{8}\text{C}_1}{\ ^{18}\text{C}_1}$
$=\frac{6\times4\times8\times3\times2}{18\times17\times16}=\frac{7}{17}$
$\therefore\text{P}\text{(E)}=\frac{4}{17}$

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Question 73 Marks

Tickets numbered form 1 to 20 are mixed up together and then a ticket is drawn at random,what is the probability that the ticket has a number which is a multiple of 3 or 7?

Answer

Since one ticket is drawn form a mixed numbers (1 to 20) tickets.
$\therefore\text{n}\text{(S)}=\ ^{20}\text{C}_1=20$
Let E be the events of getting ticket which has numbers that is multiple of 3 or 7.
$\therefore\text{E}=\big\{3, 6, 7, 9, 12, 14, 15, 18\big\} $
$\therefore\text{n}\text{(E)}=8$
$\therefore\text{P}\text{(E)}=\frac{8}{20}=\frac{2}{5}$
$\therefore\text{P}\text{(E)}=\frac{2}{5}$

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Question 83 Marks

In shuffling a pack of 52 playing cards, form four are accidently dropped; find the missing cards should be one form each suit.

Answer

Since from well-shuffled pack of cards, 4 cards missed out
$\therefore\ \text{n}\text{(E)}=\ ^{52}\text{C}_4$
Let E be the event that four missing cards are from each suit
$\therefore\text{n}\text{(E)}=\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1$
$\therefore\text{P}\text{(E)}=\frac{\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1\times\ ^{13}\text{C}_1}{\ ^{52}\text{C}_4}$
$=\frac{13\times13\times13\times13}{\frac{52\times51\times50\times49}{4\times3\times2\times1}}$
$=\frac{2197}{20825}$

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Question 93 Marks

In a race, the odds in favour of horses A, B, C, D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find probability that one of them wins the race.

Answer

We have,
$\text{P}(\text{A}):\text{P}(\overline{\text{A}})=1:3$
$\Rightarrow\text{P}(\text{A})=\frac{1}{4}$
$\text{P}(\text{B}):\text{P}(\overline{\text{B}})=1:4$
$\Rightarrow\text{P}(\text{B})=\frac{1}{5}$
$\text{P}(\text{C}):\text{P}(\overline{\text{CB}})=1:5$
$\Rightarrow\text{P}(\text{C})=\frac{1}{6}$
$\text{P}(\text{D}):\text{P}(\overline{\text{D}})=1:6$
$\Rightarrow\text{P}(\text{D})=\frac{1}{7}$
$\therefore$ Probability that atleast one of them wins is given by $\text{P}(\text{A}\cup\text{B}\cup\text{C}\cup\text{D})$
$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})+\text{P}(\text{D})$
$=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$
$=\frac{319}{420}$

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Question 103 Marks

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of its being a spade or a king.

Answer

$\because$ 1 card is drawn from a well shuffled deck of 52 cards
$\therefore\text{S}=^{52}\text{C}_1=52$
Now,
The favourable events is that drawn card is either spade or a king
Let A = Event of choosing shade
$\Rightarrow\ ^{13}\text{C}_1=13$
B = Event of choosing a king
$\Rightarrow\ ^{4}\text{C}_1=4$
Also, king can be of spade
$\therefore(\text{A}\cap\text{B})=1$
$\therefore\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}$
$=\frac{16}{52}$
$=\frac{4}{13}$

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Question 113 Marks

A box contains 10 white, 6 red and 10 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either white or red?

Answer

Let W be the event of drawing white ball
$\therefore\text{p}(\text{W})=\frac{10}{26}$
Let R be the event of drawing red ball
$\therefore\text{p}(\text{R})=\frac{6}{26}$
$\therefore\text{P}(\text{W}\cup\text{R})=\text{p}(\text{W})+\text{p}(\text{R})-\text{P}(\text{W}\cap\text{R})$ [$\because$ W and R are mutually exclusive case]
$\therefore\text{P}(\text{W}\cap\text{R})=0$
$=\frac{10}{26}+\frac{6}{26}-0$
$=\frac{16}{26}$
$=\frac{8}{13}$

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Question 123 Marks

The probability that a person will travel by plane is $\frac{3}{5}$ and that he will travel by trains is $\frac{1}{4}.$ What is the probability that he (she) will travel by plane or train?

Answer

Let A be the event that the person travel by plane
$\text{P}(\text{A})=\frac{3}{5}$
Let B be the event that the person travel by train
$\text{P}(\text{B})=\frac{1}{4}$
$\therefore$ A and B are mutually exclusive case.
$\therefore\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})$
$=\frac{3}{5}+\frac{1}{4}$
$=\frac{17}{20}$

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Question 133 Marks

A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

Answer

Let A be the event of getting an ace
$\text{P}(\text{A})=\frac{4}{52}=\frac{1}{13}$
B be the event of getting a spade card
$\text{P}(\text{B})=\frac{13}{52}=\frac{1}{4}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{52}$ [In a spade suit there is one ace also]
$\therefore\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{13}+\frac{1}{4}-\frac{1}{52}$
$=\frac{16}{52}$
$=\frac{4}{13}$

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Question 143 Marks

A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that (i) both the tickets have prime numbers on them (ii) on one there is a prime number and on the other there is a multiple of 4.

Answer

Bag has tickets numbered from 1 to 20 two tickets are drawn

$\Rightarrow\text{n}(\text{S})=^{20}\text{C}_2$

Let E be the event that both the tickets have prime number on them

$\therefore\text{n}(\text{E})=^{8}\text{C}_2=56$ [as there are 8 prime numbers between 1 to 2 as 2, 3, 5, 7, 11, 13, 17, 19]

$\therefore\text{p}(\text{E})=\frac{56}{^{20}\text{C}_2}=\frac{56}{20\times19}=\frac{14}{95}$

Let E be the event that one tickets has prime numbers and other has multiple of 4.

$\therefore\text{n}(\text{E})=8\times5=40$

$\text{p}(\text{E})=\frac{40}{^{20}\text{C}_2}=\frac{40\times2}{20\times19}=\frac{4}{19}$ [$\because$ (4, 8, 12, 16, 20) are multiples of 4]

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Question 153 Marks

100 students appeared for two examination, 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has passed at least one examination.

Answer

$\text{n}(\text{S})=100$
Let A be the event that students passed in first examination
$\therefore\text{p}(\text{A})=\frac{60}{100}$ [60 students were passed in first exam]
Let B be the event that students passed in second examination
$\therefore\text{p}(\text{B})=\frac{50}{100}$ [50 students were passed in second exam]
$\text{P}(\text{A}\cap\text{B})=\frac{30}{100}$ [$\because$ 30 students passed in both exam]
$\therefore\text{P}(\text{A}\cup\text{B})=\text{p}(\text{A})+\text{p}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{60}{100}+\frac{50}{100}-\frac{20}{100}=\frac{8}{100}$
$=\frac{4}{5}$

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Question 163 Marks

The face cards are removed frm a full pack. out of the remaining 40 cards, 4 are drawn at random. what is the probability that 6they belong to different suits?

Answer

Since face cards are removed so each suit has 10 cards each.
Now,
Four cards are drawn
$\therefore\text{n}\text{(E)}=\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1$
Let E be the event that 4 cards belongs to different suit
$\therefore\text{P}\text{(E)}=\frac{\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1\times\ ^{10}\text{C}_1}{\ ^{40}\text{C}_4}$
$=\frac{1000}{9139}$

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Question 173 Marks

From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.

Answer

Since 4 cards are drawn from a well schuffled pack of cards.
$\text{n}(\text{S})=^{52}\text{C}_4$
Let A be the event of getting 4 cards of same colour
Since there are two colours of cards
$\therefore\text{n}(\text{A})=2\ ^{26}\text{C}_4$
$\text{p}(\text{A})=2\times\frac{^{26}\text{C}_4}{^{52}\text{C}_4}$
$=2\times\frac{26\times25\times24\times23}{52\times51\times50\times49}$
$=\frac{92}{833}$

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Question 183 Marks

A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random, what is the probability that either both are rusted or both are bolts?

Answer

Box 30-bolts
40-nuts
Half the bolts and nuts are rusted
$\therefore$ rusted bolts = 15
$\therefore$ rusted bolts = 20
Since two items are drawn
$\therefore\text{n}(\text{S})=^{70}\text{C}_2$
Let A be the event of choosing rusting item
$\therefore\text{p}(\text{A})=\frac{^{35}\text{C}_2}{^{70}\text{C}_2}$
$=\frac{35\times34}{70\times69}$
Let B be the event of choosing bolts
$\therefore\text{p}(\text{B})=\frac{^{30}\text{C}_2}{^{70}\text{C}_2}$
$=\frac{30\times29}{70\times69}$
Also, $\text{n}(\text{A}\cap\text{B})=15$ [bolts that are rusted]
$\therefore\text{p}(\text{A}\cap\text{B})=\frac{^{15}\text{C}_2}{^{70}\text{C}_2}{}$
$=\frac{15\times14}{70\times69}$
$\therefore\text{p}(\text{A}\cup\text{B})=\text{p}(\text{A})+\text{p}(\text{B})-\text{p}(\text{A}\cap\text{B})$
$=\frac{35\times34}{70\times69}+\frac{30\times29}{70\times69}-\frac{15\times14}{70\times69}$
$=\frac{1850}{4830}$
$=\frac{185}{483}$

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Question 193 Marks

In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Answer

Let A be the event that choosing student who passed the first exam.
$\therefore\text{p}(\text{A})=0.8$
Let B be the event that choosing student who passed the 2nd exam.
$\text{p}(\text{B})=0.7$
$\text{n}(\text{A}\cup\text{B})=$ number of students who passed atleast one of the two exams
$\Rightarrow\text{p}(\text{A}\cup\text{B})=0.95$
$\text{p}(\text{A}\cup\text{B})=\text{p}(\text{A})+\text{p}(\text{B})-\text{p}(\text{A}\cap\text{B})$
$\text{p}(\text{A}\cap\text{B})=\text{p}(\text{A})+\text{p}(\text{B})-\text{p}(\text{A}\cup\text{B})$
$=0.8+0.7-0.95$
$=-0.95+1.5$
$=1.5-0.95$
$=0.55$

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Question 203 Marks

There are four men and six women on the city councils. if one council member is selected for a committee at random, how likel6y is that it is a women?

Answer

There are four men and six women on the city councils.
$\because$ one council member is selected for a committe.
$\therefore\text{n}\text{(S)}=\ ^{10}\text{C}_1=10$
Let E be the events that is a women,
$\therefore\text{n}\text{(E)}=\ ^{6}\text{C}_1=6$
$\therefore\text{P}\text{(E)}=\frac{6}{10}=\frac{3}{5}$

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Question 213 Marks

A dice is thrown twice. What is the probability that at least one of the two throws come up with the number 3?

Answer

A die is thrown twice
$\Rightarrow\text{n}(\text{S})=36$
Let A be the event of getting 3 in first throw
$\therefore\text{P}(\text{A})=\frac{6}{36}=\frac{1}{6}$ $\big[\because\text{A}=\big\{(3,\ 1),\ (3,\ 2),\ (3,\ 3),\ (3,\ 4),\ (3,\ 5)\big\}\big]$
B be the event of getting 3 in 2nd throw
$\text{P}(\text{B})=\frac{6}{36}=\frac{1}{6}$
Also, $\text{P}({\text{A}\cap\text{B}})=\frac{1}{36}$ $\big[\because{\text{A}\cap\text{B}}=\{3,\ 3\}\big]$
$\therefore\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}$
$=\frac{11}{36}$

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