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Maths 2 Marks Questions

P-1 Probability · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 52 questions.

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Question 12 Marks
Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card?
Answer
Probability of the selection that the card drawn is a vowel card, $p(V)=\frac{\text { Number of vowels }}{\text { Total number of alphabets }}$
$\therefore p ( V )=\frac{5}{26}$
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Question 22 Marks
In a hockey team there are 6 defenders , 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that -
The goalee will be selected.
Answer
Probability of the selection that the goalee will be selected, $p(G)=\frac{\text { Number of goalee }}{\text { Total number of players }}$
$\therefore p ( G )=\frac{1}{11}$
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Question 32 Marks
In a hockey team there are 6 defenders , 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that -
A defender will be selected.
Answer
Probability of the selection that the goalee will be selected, $p(D)=\frac{\text { Number of goalee }}{\text { Total number of players }}$
$\therefore p(D)=\frac{6}{11}$
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Question 42 Marks
Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are 4/5 0.83 and 58% respectively. Who had the greatest probability of success?
Answer
Probability of success for John= $\frac{4}{5} = 0.8$
Probability of success for Vasim = 0.83
Probability of success for Akash = 0.58
$\therefore$ Probability of success is highest for = Vasim
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Question 52 Marks
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is -
a spade.
Answer
$\text { Probability that the card drawn is a spade, } p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$ \therefore p ( B )=\frac{1}{4}$
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Question 62 Marks
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is -
an ace.
Answer
Probability that the card drawn is an ace, $p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p ( A )=\frac{1}{13}$
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Question 72 Marks
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
an odd number ?
Answer
$\text { Probability that the card drawn is an ace, } p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} $
$ \therefore p ( A )=\frac{1}{13}$
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Question 82 Marks
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
a multiple of 5 ?
Answer
Probability that the number formed is an odd number,
$p ( X )=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} $
$ \therefore p ( X )=\frac{1}{5}$
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Question 92 Marks
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
shows an even number.
Answer
Probability of event that the ticket drawn shows an even number, $p(E)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p ( E )=\frac{7}{15}$
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Question 102 Marks
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
shows a number which is a multiple of 5.
Answer
Probability of event that the ticket drawn shows a number which is a multiple of 5,
$p(X)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} $
$ \therefore p(X)=\frac{1}{5}$
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Question 112 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The digit on the first die is greater than the digit on second die.
Answer
Probability of getting the digit on the first die is greater than the digit on second die, p(C)=$\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
Favorable Outcomes are as follow:- (2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3),(6,4), (6,5)= 15 cases
$\therefore p(C)=\frac{15}{36}=\frac{5}{12}$
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Question 122 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The sum of the digits on the upper faces is 33.
Answer
Probability of getting the sum of the digits on the upper faces is $33, p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$ Favorable Outcomes(Getting the sum of Digits on the upper Faces is 33 ) $=0$ as the maximum sum could be 12
$\therefore p ( B )=0$
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Question 132 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The sum of the digits on the upper faces is at least 10.
Answer
Probability of getting the sum of the digits on the upper faces is at least 10, $p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
We know the Favourable Cases are (Where the Sum of digits on the upper faces is at least 10):- (4,6), (5,5), (5,6), (6,4), (6,5), (6,6)- 6 cases
Total Number of Outcomes:- 36
$\therefore p ( A )=\frac{6}{36}=\frac{1}{6}$
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Question 142 Marks
If two coins are tossed, find the probability of the following events.
Getting at least one head.
Answer
Sample Space, S=(HH,HT,TH,TT)
(1)Probability of getting at least one head, $p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p ( A )=\frac{3}{4}$
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Question 152 Marks
If two coins are tossed, find the probability of the following events.
Getting no head.
Answer
Probability of getting no head, $p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p ( B )=\frac{1}{4}$
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Question 162 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
One coin and one die are thrown simultaneously.
Condition for event A : To get head and an odd number.
Condition for event B : To get a head or tail and an even number.
Condition for event C : Number on the upper face is greater than 7 and tail on the coin.

Answer
Sample Space, S = {H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6}


Number of sample points, n(S) = 12


To get head and an odd number , A = {H1, H3, H5}


n(A) = 3


To get a head or tail and an even number , B = {H2, H4, H6, T2, T4, T6}


n(B) = 6


Number on the upper face is greater than 7 and tail on the coin= No Sample space exists for this event as dice does not have numbers greater than or equal to 7


n(C)=0


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Question 172 Marks
Write sample space ' $S$ ' and number of sample point $n(S)$ for each of the following experiments. Also write events $A$, $B, C$ in the set form and write $n(A), n(B), n(C)$.
From three men and two women, environment committee of two persons is to be formed.
Condition for event $A$ : There must be at least one woman member.
Condition for event B : One man, one woman committee to be formed
Condition for event $C$ : There should not be a woman member.

 
Answer
Sample Space, $S=\left\{M_1 M_2, M_1 M_3, M_1 F_1, M_1 F_2, M_2 M_3, M_2 F_1, M_2 F_2, M_3 F_1, M_3 F_2, F_1 F_2\right\}$
Number of sample points, $n(S)=10$
There must be at least one woman member, $A=\left\{M_1 F_1, M_1 F_2, M_2 F_1, M_2 F_2, M_3 F_1, M_3 F_2, F_1 F_2\right\}$
$n(A)=7$
One man, one woman committee to be formed, $B=\left\{ M _1 F_1, M _1 F_2, M _2 F_1, M _2 F_2, M _3 F_1, M _3 F_2\right\}$
$n(B)=6$
There should not be a woman member, $C=\left\{M_1 M_2, M_1 M_3, M_2 M_3\right\}$
$n(C)=3$
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Question 182 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A : The number formed is even
Condition for event B : The number formed is divisible by 3.
Condition for event C : The number formed is greater than 50.

Answer
Sample Space, S= {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43,45, 50, 51, 52, 53, 54}


Number of sample points, n(S) = 25


The number formed is even , A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}


n(A) = 13


The number formed is divisible by 3, B = {12, 15, 21, 24, 30, 42, 45, 51, 54}


n(B) = 9


The number formed is greater than 50,C = {51, 52, 53, 54} n(C) = 4


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Question 192 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Three coins are tossed simultaneously.
Condition for event A : To get at least two heads.
Condition for event B : To get no head.
Condition for event C : To get head on the second coin.

Answer
Sample Space, S = {HHH, HHT, HTT, HTH, THT, TTH, THH, TTT}


Number of sample points, n(S) = 8


To get at least two heads ,A = {HHH, HHT, HTH, THH}


n(A) = 4


To get no head, B = {TTT}


n(B) = 1


To get head on the second coin ,C = {HHH, HHT, THH}


n(C) = 3


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Question 202 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Two dice are rolled simultaneously,
Event A : The sum of the digits on upper faces is a multiple of 6.
Event B : The sum of the digits on the upper faces is minimum 10.
Event C : The same digit on both the upper faces.

Answer
Sample space, S = {(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)}


Number of sample points, n(S) = 36


A = {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (6, 6)}, n(A) = 6


B = {(4, 6) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)} , n(B) = 6


C = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)} , n(C) = 6


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Question 212 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
One die is rolled,
Event A : Even number on the upper face.
Event B : Odd number on the upper face.
Event C : Prime number on the upper face.

Answer
Sample space, S = {1, 2, 3, 4, 5, 6}


Number of sample points, n(S) = 6


A = {2, 4, 6} , n(A) = 3


B = {1, 3, 5} , n(B) = 3


C = {2, 3, 5} , n(C) = 3


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Question 222 Marks
A box contains $5$ strawberry chocolates, $6$ coffee chocolates and $2$ peppermint chocolates. Find the probability of each of the following events, if one of the chocolates is picked from the box at random. (i) it is a coffee chocolate. (ii) it is a peppermint chocolate.
Answer
Sample space is ' $S$ ' and $n(S)=5+6+2=13$
$\text { Event A : it is a coffee chocolate }$
$\therefore \quad \mathrm{n}(\mathrm{A})=6$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{6}{13}$
$\text { Event B : it is a peppermint chocolate }$
$\therefore \quad \mathrm{n}(\mathrm{B})=2$
$\therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{13}$
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Question 232 Marks
A card is drawn from a well shuffled pack of 52 playing cards. Find the probability of each event. The card drawn is (i) a red card (ii) a face card
Answer
' $\mathrm{S}$ ' is the sample space.
$\therefore \mathrm{n}(\mathrm{S})=52$
Event A : Card drawn is a red card.
Total red cards $=13$ hearts +13 diamonds $=26$
$
\begin{array}{c}
\therefore \mathrm{n}(\mathrm{A})=26 \\
\therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}=\frac{1}{2}
\end{array}
$
Event B : Card drawn is a face card.
Total face cards $=12 \quad \therefore \mathrm{n}(\mathrm{B})=12$
$
\therefore P(B)=\frac{n(B)}{n(S)}=\frac{12}{52}=\frac{3}{13}
$
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Question 242 Marks
If one die is rolled then find the probability of each of the following events.
(i) Number on the upper face is prime
(ii) Number on the upper face is even.
Answer
' $\mathrm{S}$ ' is the sample space.
$
\mathrm{S}=\{1,2,3,4,5,6\} \quad \therefore \mathrm{n}(\mathrm{S})=6
$
(i) Event A : Prime number on the upper face.
$
\begin{array}{c}
A=\{2,3,5\} \quad \therefore n(A)=3 \\
P(A)=\frac{n(A)}{n(S)} \\
\therefore P(A)=\frac{3}{6}=\frac{1}{2}
\end{array}
$
(ii) Event B : Even number on the upper face.
$
\begin{array}{c}
B=\{2,4,6\} \quad \therefore n(B)=3 \\
P(B)=\frac{n(B)}{n(S)} \\
\therefore P(B)=\frac{3}{6}=\frac{1}{2}
\end{array}
$
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Question 252 Marks
Find the probability of the following, when one coin is tossed.
(i) getting head (ii) getting tail
Answer
Let ' $S$ ' be the sample space.
$\mathrm{S}=\{\mathrm{H}, \mathrm{T}\} \quad \mathrm{n}(\mathrm{S})=2$
(i) Let event $A$ be getting head
$ A=\{H\}$
$P(A)=\frac{n(A)}{n(S)}=\frac{1}{2}$
(ii) Let event $B$ be getting tail
$B=\{T\} \quad \therefore n(B)=1$
$P(B)=\frac{n(B)}{n(S)}=\frac{1}{2}$
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Question 262 Marks
A bag contains 50 cards. Each card bears only one number from 1 to 50. One
card is drawn at randomfrom the bag. Write the sample space. Also write the
events A, B and find the number of sample points in them.
(i) Condition for event A : the number on the card is divisible by 6.
(ii) Condition for event B : the number on the card is a complete square.
Answer
S = {1, 2, 3, . . . 49, 50}, n(S) = 50
(i) Condition for event A : number is divisible by 6.
A = {6, 12, 18, 24, 30, 36, 42, 48} n(A) = 8
(ii) Condition for event B : the number on the card is a complete square.
B = {1, 4, 9, 16, 25, 36, 49} n(B) = 7
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Question 272 Marks
Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are 4/5 0.83 and 58% respectively. Who had the greatest probability of success?
Answer
Probability that the card drawn is a spade, $p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p(B)=\frac{1}{4}$
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Question 282 Marks
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is -
a spade.
Answer
Probability that the card drawn is a spade, $p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p(B)=\frac{1}{4}$
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Question 292 Marks
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is -
an ace.
Answer
Probability that the card drawn is an ace, $p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p(A)=\frac{1}{13}$
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Question 302 Marks
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
a multiple of 5 ?
Answer
Probability that the number formed is an odd number,
$\begin{array}{l} 
p ( X )=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} \\
\therefore p ( X )=\frac{1}{5}
\end{array}$
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Question 312 Marks
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
an odd number ?
Answer
Probability that the number formed is an odd number,
$\begin{array}{l}
p(X)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} \\
\therefore p(X)=\frac{4}{5}
\end{array}$
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Question 322 Marks
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
shows a number which is a multiple of 5.
Answer
Probability of event that the ticket drawn shows a number which is a multiple of 5 ,
$\begin{array}{l} 
p ( X )=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }} \\
\therefore p ( X )=\frac{1}{5}
\end{array}$
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Question 332 Marks
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
shows an even number.
Answer
Probability of event that the ticket drawn shows an even number, $p(E)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p(E)=\frac{7}{15}$
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Question 342 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The digit on the first die is greater than the digit on second die.
Answer
Probability of getting the digit on the first die is greater than the digit on second die, $p(C)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
Favorable Outcomes are as follow:- $(2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)=15$ cases
$\therefore p(C)=\frac{15}{36}=\frac{5}{12}$
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Question 352 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The sum of the digits on the upper faces is 33.
Answer
Probability of getting the sum of the digits on the upper faces is $33, p(B)=\frac{\text { favourable outcomes }}{\text { Total number of outcomes }}$ navorable Outcomes(Getting the sum of Digits on the upper Faces is 33$)=0$ as the maximum sum could be 12
$\therefore p(B)=0$
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Question 362 Marks
If two dice are rolled simultaneously, find the probability of the following events.
The sum of the digits on the upper faces is at least 10.
Answer
Probability of getting the sum of the digits on the upper faces is at least $10, p(A)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
We know the Favourable Cases are (Where the Sum of digits on the upper faces is at least 10$)$ ): $(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$ 6 cases
Total Number of Outcomes:- 36
$\therefore p(A)=\frac{6}{36}=\frac{1}{6}$
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Question 372 Marks
If two coins are tossed, find the probability of the following events.
Getting at least one head.
Answer
Sample Space, $S=(H H, H T, T H, T T)$
(1)Probability of getting at least one head, $p(A)=\frac{\text { Favourable outcome }}{\text { Fotal number of outcomes }}$
$\therefore p ( A )=\frac{3}{4}$
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Question 382 Marks
If two coins are tossed, find the probability of the following events.
Getting no head.
Answer
Probability of getting no head, $p(B)=\frac{\text { Favourable outcome }}{\text { Total number of outcomes }}$
$\therefore p(B)=\frac{1}{4}$
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Question 392 Marks
Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card?
Answer
Probability of the selection that the card drawn is a vowel card, $p(V)=\frac{\text { Number of vowels }}{\text { Total number of alphabets }}$
$\therefore p(V)=\frac{5}{26}$
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Question 402 Marks
In a hockey team there are 6 defenders , 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that -
A defender will be selected.
Answer
Probability of the selection that the goalee will be selected, $p(D)=\frac{\text { Number of goalee }}{\text { Total number of players }}$
$\therefore p(D)=\frac{6}{11}$
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Question 412 Marks
In a hockey team there are 6 defenders , 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that -
The goalee will be selected.
Answer
Probability of the selection that the goalee will be selected, $p(G)=\frac{\text { Number of goalee }}{\text { Total number of players }}$
$\therefore p(G)=\frac{1}{11}$
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Question 422 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
One coin and one die are thrown simultaneously.
Condition for event A : To get head and an odd number.
Condition for event B : To get a head or tail and an even number.
Condition for event C : Number on the upper face is greater than 7 and tail on the coin.

Answer
Sample Space, S = {H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6}
Number of sample points, n(S) = 12
To get head and an odd number , A = {H1, H3, H5}
n(A) = 3
To get a head or tail and an even number , B = {H2, H4, H6, T2, T4, T6}
n(B) = 6
Number on the upper face is greater than 7 and tail on the coin= No Sample space exists for this event as dice does not have numbers greater than or equal to 7
n(C)=0
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Question 432 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
From three men and two women, environment committee of two persons is to be formed.
Condition for event A : There must be at least one woman member.
Condition for event B : One man, one woman committee to be formed
Condition for event C : There should not be a woman member.

Answer
Sample Space,
S = {M1M2, M1M3, M1F1, M1F2, M2M3, M2F1, M2F2, M3F1, M3F2, F1F2}
Number of sample points, n(S) = 10
There must be at least one woman member ,
A = {M1F1, M1F2, M2F1, M2F2, M3F1, M3F2, F1F2}
n(A) = 7
One man, one woman committee to be formed ,
B = {M1F1, M1F2, M2F1, M2F2, M3F1, M3F2}
n(B) = 6
There should not be a woman member ,
C = {M1M2, M1M3, M2M3}
n(C) = 3
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Question 442 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A : The number formed is even
Condition for event B : The number formed is divisible by 3.
Condition for event C : The number formed is greater than 50.
Answer
Sample Space, S= {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43,45, 50, 51, 52, 53, 54}
Number of sample points, n(S) = 25
The number formed is even , A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}
n(A) = 13
The number formed is divisible by 3, B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
n(B) = 9
The number formed is greater than 50,C = {51, 52, 53, 54} n(C) = 4
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Question 452 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Three coins are tossed simultaneously.
Condition for event A : To get at least two heads.
Condition for event B : To get no head.
Condition for event C : To get head on the second coin.
Answer
Sample Space, S = {HHH, HHT, HTT, HTH, THT, TTH, THH, TTT}
Number of sample points, n(S) = 8
To get at least two heads ,A = {HHH, HHT, HTH, THH}
n(A) = 4
To get no head, B = {TTT}
n(B) = 1
To get head on the second coin ,C = {HHH, HHT, THH}
n(C) = 3
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Question 462 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
Two dice are rolled simultaneously,
Event A : The sum of the digits on upper faces is a multiple of 6.
Event B : The sum of the digits on the upper faces is minimum 10.
Event C : The same digit on both the upper faces.

Answer
Sample space, S = {(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)}
Number of sample points, n(S) = 36
A = {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (6, 6)}, n(A) = 6
B = {(4, 6) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)} , n(B) = 6
C = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)} , n(C) = 6
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Question 472 Marks
Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
One die is rolled,
Event A : Even number on the upper face.
Event B : Odd number on the upper face.
Event C : Prime number on the upper face.

Answer
Sample space, S = {1, 2, 3, 4, 5, 6}
Number of sample points, n(S) = 6
A = {2, 4, 6} , n(A) = 3
B = {1, 3, 5} , n(B) = 3
C = {2, 3, 5} , n(C) = 3
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Question 482 Marks
A card is drawn from a well shuffled pack of 52 playing cards. Find the probability of each event. The card drawn is (i) a red card (ii) a face card
Answer
$\mathrm{S}$ ' is the sample space.
$\therefore \mathrm{n}(\mathrm{S})=52$
Event A : Card drawn is a red card.
Total red cards $=13$ hearts +13 diamonds $=26$
$
\begin{array}{c}
\therefore \mathrm{n}(\mathrm{A})=26 \\
\therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}=\frac{1}{2}
\end{array}
$
Event B : Card drawn is a face card.
Total face cards $=12 \quad \therefore \mathrm{n}(\mathrm{B})=12$
$
\therefore P(B)=\frac{n(B)}{n(S)}=\frac{12}{52}=\frac{3}{13}
$
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Question 492 Marks
A box contains 5 strawberry chocolates, 6 coffee chocolates and 2 peppermint chocolates. Find the probability of each of the following events, if one of the chocolates is picked from the box at random. (i) it is a coffee chocolate. (ii) it is a peppermint chocolate.
Answer
Sample space is ' $S$ ' and $n(S)=5+6+2=13$
$
\begin{array}{l}
\text { Event A : it is a coffee chocolate } \\
\therefore \quad \mathrm{n}(\mathrm{A})=6 \\
\therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{6}{13} \\
\text { Event B : it is a peppermint chocolate } \\
\therefore \quad \mathrm{n}(\mathrm{B})=2 \\
\therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{13} \\
\end{array}
$
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Question 502 Marks
If one die is rolled then find the probability of each of the following events.
(i) Number on the upper face is prime
(ii) Number on the upper face is even.
Answer
$\mathrm{S}$ ' is the sample space.
$
\mathrm{S}=\{1,2,3,4,5,6\} \quad \therefore \mathrm{n}(\mathrm{S})=6
$
(i) Event A : Prime number on the upper face.
$
\begin{array}{c}
A=\{2,3,5\} \quad \therefore n(A)=3 \\
P(A)=\frac{n(A)}{n(S)} \\
\therefore P(A)=\frac{3}{6}=\frac{1}{2}
\end{array}
$
(ii) Event B : Even number on the upper face.
$
\begin{array}{c}
B=\{2,4,6\} \quad \therefore n(B)=3 \\
P(B)=\frac{n(B)}{n(S)} \\
\therefore P(B)=\frac{3}{6}=\frac{1}{2}
\end{array}
$
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Question 512 Marks
Find the probability of the following, when one coin is tossed.
(i) getting head (ii) getting tail
Answer
Let ' $S$ ' be the sample space.
$\mathrm{S}=\{\mathrm{H}, \mathrm{T}\} \quad \mathrm{n}(\mathrm{S})=2$
(i) Let event $A$ be getting head
$\begin{array}{l}
A=\{H\} \
P(A)=\frac{n(A)}{n(S)}=\frac{1}{2}
\end{array}$
(ii) Let event $B$ be getting tail
$\begin{array}{l}
B=\{T\} \quad \therefore n(B)=1 \
P(B)=\frac{n(B)}{n(S)}=\frac{1}{2}
\end{array}$ 
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Question 522 Marks
A bag contains 50 cards. Each card bears only one number from 1 to 50. One
card is drawn at randomfrom the bag. Write the sample space. Also write the
events A, B and find the number of sample points in them.
(i) Condition for event A : the number on the card is divisible by 6.
(ii) Condition for event B : the number on the card is a complete square.
Answer
S = {1, 2, 3, . . . 49, 50}, n(S) = 50
(i) Condition for event A : number is divisible by 6.
A = {6, 12, 18, 24, 30, 36, 42, 48} n(A) = 8
(ii) Condition for event B : the number on the card is a complete square.
B = {1, 4, 9, 16, 25, 36, 49} n(B) = 7
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