2 Marks Questions - Maths STD 10 Questions - Vidyadip

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2 Marks Questions

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

Question 12 Marks

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure) and these are equally likely outcomes. What is the probability that it will point at:

8?
an odd number?
a number greater than 2?
a number less than 9?

Answer

Probability that two students are not having same birthday = P(E) = 0.992
$$We know, probability of occurrence of an event and probability of non occurrence of event = 1
$\therefore$ P(E) + $P ( \overline { E } )$ = 1
$\Rightarrow$0.992 + $P ( \overline { E } )$ = 1
$\Rightarrow$$P ( \overline { E } )$ = 1 - 0.992
$\Rightarrow$$P ( \overline { E } )$ = 0.008
Hence, P(two students have the same birthday) = 0.008

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Question 22 Marks

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

an orange flavoured candy?
a lemon flavoured candy?

Answer

Since, P(E) + P (not E) = 1

P (not E) = 1 - P(E) = 1 - 0.05 = 0.95

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Question 32 Marks

Is the given statement correct or not correct?
If a die is thrown, there are two possible outcomes- an odd number or an even number. Therefore, the probability of getting an odd number is $\frac{1}{2}$.

Answer

Total outcomes that can occur are 1, 2, 3, 4, 5, 6
Number of possible outcomes of a dice = 6
Numbers which are odd = 1, 3, 5
Total numbers which are odd = 3
Numbers which are even = 2, 4, 6
Total numbers which are even = 3
Probability of getting an odd number $=\frac{\text {Number of outcomes where there is an odd number}}{\text {Total number of outcomes}}$
$=\frac{3}{6}=\frac{1}{2}$
Hence, the given statement is correct.

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Question 42 Marks

Is the given statement correct or not correct?
If two coins are tossed simultaneously there are three possible outcomes- two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac{1}{3}$.

Answer

All possible outcomes = (H, H), (H, T), (T, H), (T, T)
Probability of an event = $\frac{\text { Favourabe outcomes }}{\text { Total outcomes }}$
Probability (2 heads) = $\frac{1}{4}$
Probability (2 tails) = $\frac{1}{4}$
It is not correct.
If we want to get the probability of them we should categorize the outcomes like this but they are not equally likely because one of each can result in two ways from a head-on first coin and tail on second or from the tail on first and head on second.

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Question 52 Marks

A die is thrown twice. What is the probability that:

5 will not come up either time?
5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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Question 62 Marks

A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

Since a coin is tossed 3 times
$\therefore$ Possible outcomes are = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Total possible outcomes = 8
outcomes having 3 heads and 3 tails {HHH,TTT}

$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$

P(Hanif will win the game) = $\frac28=\frac14$

P (Hanif will lose the game) =$\frac11-\frac14=\frac{4-1}4=\frac34$

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Question 72 Marks

Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

Event: Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12

Probability $\frac{1}{{36}}$ $\frac{5}{{36}}$ $\frac{1}{{36}}$
A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $\frac{1}{11}$. Do you agree with this argument? Justify your answer.

Answer

It can be observed that,
To get the sum as 2, possible outcomes = (1, 1)
To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2), (3, 3)
To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)
To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3), (4, 4)
To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6), (6, 5)
To get the sum as 12, possible outcomes = (6, 6)

Event	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{{36}}$	$\frac{2}{{36}}$	$\frac{3}{{36}}$	$\frac{4}{{36}}$	$\frac{5}{{36}}$	$\frac{6}{{36}}$	$\frac{5}{{36}}$	$\frac{4}{{36}}$	$\frac{3}{{36}}$	$\frac{2}{{36}}$	$\frac{1}{{36}}$

The probability of each of these sums will not be $\frac{1}{11}$ as their sums are not equally likely.

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Question 82 Marks

A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

she will buy it?
she will not buy it?

Answer

Total number of favourable outcomes = 144

$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$

Number of non-defective pens = 144 - 20 = 124
Number of favourable outcomes = 124
Hence P (she will buy) = P (a non-defective pen) = $\frac{124}{144}=\frac{31}{36}$
Number of favourable outcomes = 20
Hence P (she will not buy) = P (a defective pen) = $\frac{20}{144}=\frac5{36}$

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Question 92 Marks

A driver attempts to start a car. The car starts or does not start, is this experiment has an equally likely outcome? Explain.

Answer

0, 1

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Question 102 Marks

Suppose you drop a die at random on the rectangular region shown in Figure. What is the probability that it will land inside the circle with diameter 1m?

Answer

Total area of the given figure (rectangle) = $3\times2=6m^2$
d =1
$r=\frac12$
And Area of circle = $\mathrm{πr}^2$= $\mathrm\pi\left(\frac12\right)^2=\frac{\mathrm\pi}4\mathrm m^2$
$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$
Hence, P(die to land inside the circle) = $\frac{\frac{\mathrm\pi}4}6=\frac{\mathrm\pi}{24}$
Hence the probability of die to land inside the circle is $\frac{\mathrm\pi}{24}$

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Question 112 Marks

A child has a die whose 6 faces show the letters given below:

The die is thrown once. What is the probability of getting (i) A, (ii) D?

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Question 122 Marks

A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer

Total number of favourable outcomes = 20
$Probablity\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$
Number of favourable outcomes = 4
Hence P (getting a defective bulb) = $\frac4{20}=\frac15$
Now total number of favourable outcomes = 20 - 1 = 19
Number of favouroable outcomes = 19 - 4 = 15
Hence P (getting a non-defective bulb) = $\frac{15}{19}$

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Question 132 Marks

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

Total number of pens = 132 + 12 = 144
Number of good pens = 132
Let E be the event of getting a good pen.
Therefore, P(getting a good pen), P(E) =
$\frac { \text { Number of outcomes favorable to } E } { \text { Number of all possible outcomes } } = \frac { 132 } { 144 } = \frac { 11 } { 12 }$

Thus, the probability of getting a good pen is $\frac{{11}}{{12}}$.

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Question 142 Marks

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

What is the probability that the card is the queen?
If the queen is drawn and put aside, what is the probability that the second card picked up is
1. an ace?
2. a queen?

Answer

Total number of favourable outcomes = 5

$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$

(i) There is only one queen.
$\therefore$ Favourable outcome = 1
Hence, P (the queen) = $\frac15$
In this situation, total number of favourable outcomes = 4
1. Favourable outcome = 1
  Hence, P (an ace) = $\frac14$
2. There is no card as queen.
  $\therefore$Favourable outcome = 0
  Hence, P (the queen) = $\frac04=0$

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Question 152 Marks

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Answer

There are 13 (8 + 5) fish out of which one can be chosen in 13 ways.
Total number of elementary events = 13
There are 5 male fish out of which one male fish can be chosen in 5 ways.
Favourable number of elementary events = 5
Hence, required probability $= \frac { 5 } { 13 }$

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Question 162 Marks

There are 40 students in class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of

a girl?
a boy?

Answer

Since there are 40 students and there is one card for each student. So, one, card can be chosen out of 40 cards in 40 ways.
Total number of elementary events = 40

There are 25 girls and corresponding to each girl there is card of her name. Therefore,
a card with the name of a girl can be chosen in 25 ways.
Favourable number of elementary events = 25
Hence, P (Getting a card with the name of a girl)$= \frac { 25 } { 40 } = \frac { 5 } { 8 }$
We have,
P (Getting a card with name of a boy) = 1 - P (Getting a card with name of a girl) $= 1 - \frac { 5 } { 8 } = \frac { 3 } { 8 }$

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Question 172 Marks

Savita and Hamida are friends. What is the probability that both will have

different birthdays?
the same birthday? (ignoring a leap year)

Answer

Savita may have any one of the 365 days of the year as her birthday.
Similarly, Hamida may have any one of 365 days of the year as her birthday.
$\therefore$ Total number of ways in which Savita and Hamida may have their birthday = 365 $\times$ 365

We have,
Probability that Savita and Hamida will have different birthdays = 1
Therefore,ProbabilityProbability that Savita and Hamida will have the same birthday $= 1 - \frac { 1 } { 365 } = \frac { 364 } { 365 }$
Savita and Hamida may have same birthday on any one of 365 days of the year.
Number of ways in which Savita and Hamida will have same birthday = 365
Therefore,Probability that Savita and Hamida will have the same birthday $= \frac { 365 } { 365 \times 365 } = \frac { 1 } { 365 }$

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Question 182 Marks

Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?

Answer

Probability of winning the match by Reshma
= Probability of loosing the match by Sangeeta
= 1 - Probability of winning the match by Sangeeta
= 1 - 0.62 = 0.38

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Question 192 Marks

One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will

be an ace,
not be an ace.

Answer

Well-shuffling ensures equally likely outcomes.

There are 4 aces in a deck. Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52
Therefore, P(E) = $\frac{4}{52}=\frac{1}{13}$
Let F be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event F = 52 - 4 = 48
The number of possible outcomes = 52
Therefore, P(F) = $\frac{48}{52}=\frac{12}{13}$

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Question 202 Marks

Suppose we throw a die once.

What is the probability of getting a number greater than 4?
What is the probability of getting a number less than or equal to 4?

Answer

Let E be the event that ‘getting a number greater than 4’.
The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6.
Therefore, the number of outcomes favourable to E is 2.
so, P(E) = P(number greater than 4) = $\frac{2}{6}=\frac{1}{3}$
Let F be the event that ‘getting a number less than or equal to 4’.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4.
so, the number of outcomes favourable to F is 4.
Therefore, P(F) $=\frac{4}{6}=\frac{2}{3}$

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Question 212 Marks

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that

it is acceptable to Jimmy?
it is acceptable to Sujatha?

Answer

One shirt is drawn at random from the carton of 100 shirts. This can be done in 100 ways.
Total number of elementary events = 100.

Since Jimmy accepts only good shirts and the number of good shirts is 88
Number of elementary events favourable to Jimmy = 88
So, probability that a shirt is acceptable to Jimmy = = 0.88
Sujatha accepts good as well as shirts having minor defects.
The number of such shirts is 88 + 8 = 96
Number of elementary events favourable to an event of selecting a good shirt or a shirt with minor defect is 96.
Hence, probability that a shirt is acceptable to Sujatha$= \frac { 96 } { 100 } = 0.96$

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Question 222 Marks

A missing helicopter is reported to have crashed somewhere in the rectangular region in Fig. What is the probability that it is crashed inside the lake shown in the figure?

Answer

According to the question,
Area of the entire region where the helicopter can crash $= (4.5\times 9) km^2 = 40.5\ km^2$
Area of the lake $= (2.5\times 3) km^2 = 7.5\ km^2$
Probability that the helicopter crashed inside the lake$= \frac { 7.5 } { 40.5 } = \frac { 75 } { 405 } = \frac { 5 } { 27 }$

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