2 Marks Questions - Maths STD 10 Questions - Vidyadip

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16 questions · timed · auto-graded

Question 12 Marks

A bag contains $3$ red balls and $5$ black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

red ?
not red ?

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

Which of the following cannot be the probability of an event?

Answer

$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$
Probability of head= $P(H)$=$\frac12$
Probability of tail = $P(T)$=$\frac12$
i.e. $P\left(H\right)=\frac12=P\left(T\right)=\frac12$
Probability of getting head and tail both are same.
$\therefore$Tossing a coin considered to be fairway.

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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]

Answer

(i) The outcomes associated with the experiment in which a dice is thrown is twice:
$(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)$
$(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$
$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)$
$(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$
$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)$
$(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$
Therefore, Total number of favourable outcomes $= 36$
The number of outcomes where $5$ come at least once are $11$ Hence the number of outcomes where $5$ will not come up either time $36-11=25$
$\therefore$ $P(E)=$$P\left(5\;will\;not\;come\;up\;either\;time\right)$$=\frac{Number\;of\;favorable\;outcomes\;}{Number\;of\;total\;outcomes}$ = $\frac{25}{36}$ and $P(B)=$$P\left(5\;will\;come\;up\;at\;least\;once\right)$$=\frac{Number\;of\;favorable\;outcomes\;}{Number\;of\;total\;outcomes}$ = $\frac{11}{36}$

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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

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

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

Answer

There are $6$ letters in a die consisting of $3A's, 1B, 1C$ and $1D$.

let $E_1$ be the event of getting $A$.
number of favourable outcomes $= 3$
P(getting A)$=P(E_1)= \frac { 3 } { 6 } = \frac { 1 } { 2 }$
let $E_2$ be the event of getting $D$.
number of favourable outcomes $= 1$
$P$(getting $D$)$=P(E_2)= \frac { 1 } { 6 }$

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Question 82 Marks

A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears

a two-digit number
a perfect square number
a number divisible by $5$.

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Question 92 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 102 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?

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Question 112 Marks

A box contains $3$ blue, $2$ white, and $4$ red marbles. If a marble is drawn at random from the box, what is the probability that it will be

white?
blue?
red?

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Question 122 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 132 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 142 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 152 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?

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Question 162 Marks

In a musical chair game, the person playing the music has been advised to stop playing the music at any time within $2$ minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?

Answer

The possible outcomes are all the numbers between $0\ and\ 2$.
Suppose $A$ be the event 'music is stopped within the first half minute'.
$\therefore$ Outcomes favourable to the event $A$ are all points on the number line from $O$ to $Q$ i.e. from $0\ to\ \frac{1}{2}$

Total number of outcomes are the points on the number line from $O$ to $P$ i.e. from $0$ to $2$.
$\therefore$$P ( A ) = \frac { \text { Length } O Q } { \text { Length } O P } = \frac { 1 / 2 } { 2 } = \frac { 1 } { 4 }$

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2 Marks Questions - Maths STD 10 Questions - Vidyadip