When a pair of coins is tossed, the sample space of the experiment $S=\{ HH , HT , TH , TT \}$. Then, Which of the following probabilities is/are correct? 1. The probability of getting exactly one head is $\frac{1}{2}$. 2. The probability of getting exactly two heads is $\frac{1}{4}$. 3. The probability of getting atleast one head is $\frac{3}{4}$.
A
Only 1 and 2 are correct
B
Only 2 and 3 are correct
C
Only 3 and 1 are correct
✓
All 1, 2 and 3 are correct
Answer
Correct option: D.
All 1, 2 and 3 are correct
(d) All 1, 2 and 3 are correct Explanation: We have, $S=\{ HH , HT , TH , TT \}$ Getting exactly one head: $\{ HH , TH \}$ Probability $=\frac{2}{4}=\frac{1}{2}$ Getting exactly two heads: $\{ H \}$ Probability $=\frac{1}{4}$ Getting atleast one head: $\{ HH , HT , TH \}$ Probability $=\frac{3}{4}$
When a die is thrown, we can get any number $1,2,3,4,5,6$. So the sample space $S=\{1,2,3,4,5,6\}$. Then, Which of the following probabilities is/are correct? 1. The probability of getting an even number is $\frac{1}{2}$. 2. The probability of getting a prime number is $\frac{1}{2}$. 3. The probability of getting a number less than 5 is $\frac{1}{2}$.
✓
Only 1 and 2 are correct
B
Only 2 and 3 are correct
C
Only 3 and 1 are correct
D
All 1, 2 and 3 are correct
Answer
Correct option: A.
Only 1 and 2 are correct
(a) Only 1 and 2 are correct Explanation : We have, $S=\{1,2,3,4,5,6\}$ Getting an even number $=\{2,4,6\}$ Probability $=\frac{3}{6}=\frac{1}{2}$ Getting a prime number $=\{2,3,5\}$ Probability $=\frac{3}{6}-\frac{1}{2}$ Getting a number less than $5:\{1,2,3,4\}$ Probability $=\frac{4}{6}-\frac{2}{3}$
Statement A: If a box contains 5 white, 2 Red and 4 black marbles, then the probability of not drawing white marble from the box is $5 / 11$. Statement B: $p ( E )=1- P ( E )$, where E is any event. which of the statement are valid?
A
only A
✓
only B
C
Both A and B
D
Neither A nor B
Answer
Correct option: B.
only B
(b) Only B Explanation: Statement A: E = drawing a white marble $\begin{array}{l}n(E)=5 \\ n(S)=5+2+4=11 \\ P(E)=\frac{n(E)}{n(S)}=\frac{5}{11}\end{array}$ $P(\operatorname{not} E)=1-P(E)$ $P(E)=1-\frac{5}{11}=\frac{6}{11}$ Statement A is incorrect Statement B : $P ( E )=1- P ( E )$ is true. So, statement B is correct.
Statement A: The probability of getting a number 8 in a single throw of die. Statement B : The probability of getting a number less than 7 in a single throw of die. Which of the above statement has probability equal to 1 ?
A
only A
✓
only B
C
Both A and B
D
Neither A nor B
Answer
Correct option: B.
only B
(b) only B Explanation: For Statement $A : A =$ Getting a number 8 in a single throw of die. $\begin{array}{l}n(A)=0, n(S)=6 \\ P(A)=\frac{n(A)}{n(S)}=\frac{0}{6}=0\end{array}$ For statement $B: B=$ getting a number less than 7 in a single throw of die $n(B)=6, n(S)=6$ $P(B)=\frac{n(B)}{n(S)}=\frac{6}{6}=1$ Only statement $B$ has probability equal to 1.
Event A : The sun will rise from east tomorrow. Event B: It will rain on Monday. Event C: February month has 29 days in a leap year. Which of the above event(s) has probability equal to 1 ?
A
all events A, B and C
B
both events A and B
C
both events B and C
✓
both events A and C
Answer
Correct option: D.
both events A and C
(d) both events A and C Explanation: The probability of one means the event will definitelly occur. Chances of the event A and event C happening is $100 \%$
A school has five houses, A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single students is selected at random to be the class monitor. The probability that the selected student is not from $A , B$ and C is:
A
$\frac{4}{23}$
✓
$\frac{6}{23}$
C
$\frac{8}{23}$
D
$\frac{17}{23}$
Answer
Correct option: B.
$\frac{6}{23}$
(b) $\frac{6}{23}$ Explanation: $T(E)=23$ $F ( E )=\operatorname{not}$ from $A , B , C$, i.e., $=23-(4+8+5)$ $\begin{array}{l}F(E)=23-17=6 \\ P(F)=\frac{6}{23}\end{array}$
One ticket is drawn at random from a bag containing tickets numbered 1 to 40 . The probability that the selected ticket has a number which is a multiple of 5 is:
✓
$\frac{1}{5}$
B
$\frac{3}{5}$
C
$\frac{4}{5}$
D
$\frac{1}{3}$
Answer
Correct option: A.
$\frac{1}{5}$
(a) $\frac{1}{5}$ Explanation: $T(E)=40$ Number of outcomes favourable for event E are 5, 10, 15, 20, 25, 30, 35, 40 i.e., $F(E)=8$ $P(F)=\frac{F(F)}{T(E)}=\frac{8}{40}=\frac{1}{5}$
The probability of getting a bad egg in a lot of 400 is 0.035 . The number of bad eggs in the lot is:
A
7
✓
14
C
21
D
28
Answer
Correct option: B.
14
(b) 14 Explanation: $T(E)=400$ Number of outcomes favourable for event E , i.e., $F ( E )=$ ? $\begin{array}{l}P(F)=0.035 \\ \therefore P(F)=\frac{F(F)}{T(E)} \Rightarrow 0.035=\frac{F(E)}{400}\end{array}$ So, $F(E)=0.035 \times 400=14$ eggs. So, the number of bad eggs are 14.
An event is very unlikely to happen. Its probability is closest to:
A
0.0001
B
0.001
✓
0.01
D
$0.1$
Answer
Correct option: C.
0.01
(c) 0.0001 Explanation: The probability of the event which is very unliked to happen will be very close to zero. So it probability is 0.0001 which is minimum among the given values.
A card is drawn from a well-shuffled pack of 52 cards. The probability that the card is not an ace is:
A
$\frac{1}{13}$
B
$\frac{9}{13}$
C
$\frac{4}{13}$
✓
$\frac{12}{13}$
Answer
Correct option: D.
$\frac{12}{13}$
(d) $\frac{12}{13}$ Explanation : Total number of possible outcomes $=52$ Number of aces in the pack $=4$ Thus, the probability of not drawing an ace $\therefore P(E)=\frac{52-4}{52}=\frac{48}{52}=\frac{12}{13}$
A box contains 250 bulbs. Out of these 125 bulbs are from 'A' category, 75 bulbs are from 'B' category and the rest one are from 'C' category. The probability for 'C' category while a bulb is taken out at random is:
✓
$\frac{1}{5}$
B
$\frac{1}{7}$
C
$\frac{1}{9}$
D
$\frac{1}{10}$
Answer
Correct option: A.
$\frac{1}{5}$
(a) $\frac{1}{5}$ Explanation: Here, $n(S)=250$ and $n( E )=250-(125+75)=50$ $\therefore P(E)=\frac{n( E )}{n(S)}=\frac{50}{250}=\frac{1}{5}$.
A card is drawn at random from a well-shuffled deck of 52 cards. The probability of drawing a red or a black jack is:
✓
$\frac{1}{13}$
B
$\frac{5}{13}$
C
$\frac{1}{4}$
D
$\frac{1}{52}$
Answer
Correct option: A.
$\frac{1}{13}$
(a) $\frac{1}{13}$ Explanation: Total no. of possible outcomes, $n(S)=52$ and, total no. of favourable out comes, $n(E)=2+2=4$ $\therefore P(E)=\frac{n(E)}{n(S)}=\frac{4}{52}=\frac{1}{13} .$
A fruit box contains 120 apples. Out of these, 30 apples are rotten. The probability that the apple drawn is not rotten while an apple is drawn at random is:
A
$\frac{1}{4}$
B
$\frac{1}{2}$
✓
$\frac{3}{4}$
D
1
Answer
Correct option: C.
$\frac{3}{4}$
(c) $\frac{3}{4}$ Explanation: Here, the total no. of possible outcomes, $n(S)=120$ and, the total no. of fresh apples, $n(E)=120-30=90$ $\therefore P(E)=\frac{n(E)}{n(S)}=\frac{90}{120}=\frac{3}{4} \text {. }$
A basket contains 25 red, 15 orange and 10 purple flowers. A flower is taken out at random. The probability of selecting a purple flower from the basket is:
A
$\frac{1}{50}$
✓
$\frac{1}{5}$
C
$\frac{1}{10}$
D
$\frac{1}{100}$
Answer
Correct option: B.
$\frac{1}{5}$
(b) $\frac{1}{5}$ Explanation: Total no. of possible outcomes $\begin{array}{l}=25+15+10 \\ n(S)=50\end{array}$ $\therefore$ Total no. of favourable outcomes $\begin{array}{l}=10 \\ n(E)=10 \\ \therefore P(E)=\frac{n(E)}{n(S)}=\frac{10}{50}=\frac{1}{5} .\end{array}$
A coin is tossed 50 times and it was found that tail appear 20 times and head appears 30 times. If a coin is tossed at random, then probability of getting a tail will be:
A
$\frac{2}{5}$
B
$\frac{3}{5}$
C
1
D
None
Answer
(a) $\frac{2}{5}$ Explanation: Here, $n(S)=50$ and $n( E )=20$
A leather box contains 50 pebbles bearing numbers $1,2,3,4, \ldots \ldots \ldots, 50$ respectively. A pebble is drawn at random. The probability that the number on the pebble is a perfect square is:
✓
$\frac{7}{50}$
B
$\frac{3}{25}$
C
$\frac{1}{10}$
D
$\frac{2}{25}$
Answer
Correct option: A.
$\frac{7}{50}$
(a) $\frac{7}{50}$ Explanation: The total no. of sample space. $n(S)=50$ and the total no of favourable outcomes, $n(E)=\{1,4,9,16,25,36,49\} \text { i.e., } 7$ Now, $P(E)=\frac{n(E)}{n(S)}=\frac{7}{50}$.
A bag contains 10 coins having numbers $1,2,3,4, \ldots \ldots, 10$ respectively. A coin is drawn at random from the bag. The probability that the number on the coin is a number divisible by 2 is:
A fruit box contains 8 mangoes, 4 apples and 3 melons. One fruit is drawn at random. The probability that the fruit drawn is melon or apple will be:
A
$\frac{4}{15}$
✓
$\frac{7}{15}$
C
$\frac{3}{15}$
D
None of these
Answer
Correct option: B.
$\frac{7}{15}$
(b) $\frac{7}{15}$ Explanation: Let E be the event of getting a fruit, $\because$ Sample space $( S )=8+4+3$ $=15$ $\therefore n(S)=15$ and total no. of favourable outcomes $=4+3=7$ $\therefore n(E)=7$ Now, $P(E)=\frac{n(E)}{n(S)}=\frac{7}{15}$.
A rectangular box contains 20 cards having numbers from 1 to 20 respectively. A card is drawn at random. The probability that the number on the card is a prime number is:
✓
$\frac{2}{5}$
B
$\frac{3}{5}$
C
$\frac{4}{5}$
D
1
Answer
Correct option: A.
$\frac{2}{5}$
(a) $\frac{2}{5}$ Explanation: Sample space, $n(S)=20$ and favourable outcomes $( E )=\{2,3,5,7,11,13,17,19\}$ $\therefore n( E )=8$ $\therefore P ( E )=\frac{n( E )}{n(S)}=\frac{8}{20}=\frac{2}{5}$.
The probability that the ball drawn is red, if there are 12 red, 10 white balls in a bag and only one ball is drawn at random will be:
A
$\frac{5}{10}$
B
$\frac{5}{11}$
✓
$\frac{6}{11}$
D
$\frac{6}{7}$
Answer
Correct option: C.
$\frac{6}{11}$
(c) $\frac{6}{11}$ Explanation: The total no. of sample space. $n(S)=12+10=22$ and, the total no. of favourable outcomes, $\begin{array}{l}n(E)=12 \\ \therefore P(E)=\frac{m(E)}{n(S)}=\frac{12}{22}=\frac{6}{11}\end{array}$
One ticket is drawn at random from a bag containing tickets numbered 1 to 40 . The probability that the selected ticket has a number that is a multiple of 7 is:
A
$\frac{1}{7}$
✓
$\frac{1}{8}$
C
$\frac{1}{5}$
D
$\frac{7}{40}$
Answer
Correct option: B.
$\frac{1}{8}$
(b) $\frac{1}{8}$ Explanation: The total number of possible outcomes $=40$ The favourable outcomes are $7,14,21,28,35$. Hence, the number of favourable outcomes $=5$ Thus, $P(E)=\frac{5}{40}=\frac{1}{8}$
A coin is tossed 100 times and it was found that head appear 36 times and tail appears 64 times. If a coin is tossed at random, then probability of getting a head is:
✓
$\frac{9}{25}$
B
$\frac{5}{25}$
C
$\frac{7}{23}$
D
$\frac{6}{25}$
Answer
Correct option: A.
$\frac{9}{25}$
(a) $\frac{9}{25}$ Explanation: Number of sample space $=100$ $n(S)=100$ and, number of favourable outcomes $=36$ $n(E)=36$ $\therefore$ Probability of getting Head $\frac{n( E )}{n(S)}$ $=\frac{36}{100}=\frac{9}{23}$.
A box contains 5 white, 3 black and 7 red balls. A ball is drawn from the box at random. The probability to get red ball is:
✓
$\frac{7}{15}$
B
$\frac{8}{15}$
C
$\frac{10}{15}$
D
$\frac{12}{15}$
Answer
Correct option: A.
$\frac{7}{15}$
(a) $\frac{7}{15}$ Explanatio $\begin{array}{l}S=5+3+7=15 \\ \therefore n(S)=15\end{array}$ $E =$ getting a red ball $n(E)=7$ So, $P(E)=\frac{\pi[( E )}{n(S)}=\frac{7}{15}$