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Maths M.C.Q

Probability · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 16 questions.

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M.C.Q

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16 questions · auto-graded multiple-choice test.

MCQ 11 Mark
$80$ bulbs are selected at random from a lot and their lifetime in hours is recorded as under.
Lifetime (in hours)
$300$
$500$
$700$
$900$
$1100$
Frequency
$10$
$12$
$23$
$25$
$10$
One bulb is selected at random from the lot. What is the probability that the selected bulb has a life more than $500$ hours$?$
  • A
    $\frac{27}{40}$
  • ✓
    $\frac{29}{40}$
  • C
    $\frac{5}{16}$
  • D
    $\frac{11}{40}$
Answer
Correct option: B.
$\frac{29}{40}$
Total number of bulbs $= 80$
Probability that selected bulb has a life more than 500 hours $=\frac{23+25+10}{80}$
$=\frac{58}{80}$
$=\frac{29}{40}$
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MCQ 21 Mark
Two coins are tossed simultaneously $600$ times to get: $2$ heads : $234$ times, $1$ head : $206$ times, $0$ head : $160$ times. If two coins are tossed at random, what is the probability of getting at least one head$?$
  • A
    $\frac{103}{300}$
  • B
    $\frac{39}{100}$
  • ✓
    $\frac{11}{15}$
  • D
    $\frac{4}{15}$
Answer
Correct option: C.
$\frac{11}{15}$
Total number of outcomes $= 600$
Probability of getting at least $1$ head $=$ Probability of getting $1$ head $+$ Probability of getting $2$ heads
$=\frac{206}{600}+\frac{234}{600}$
$=\frac{440}{600}$
$=\frac{11}{15}$
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MCQ 31 Mark
In a survey of $364$ children aged $19 - 36$ months, it was found that $91$ liked to eat potato chips. If a child is selected at random, the probability that he / she does not like to eat potato chips is:
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • ✓
    $\frac{3}{4}$
  • D
    $\frac{4}{5}$
Answer
Correct option: C.
$\frac{3}{4}$
Total number of children $= 364$
Number of children who like to eat potato chips $= 91$
$⇒$ Number of children who do not like to eat potato chips $= 364 - 91 = 273$
$\therefore$ Required probability $=\frac{273}{364}=\frac{3}{4}$
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MCQ 41 Mark
In a medical examination of students of a class, the following blood groups are recorded:
Blood group
$A$
$B$
$AB$
$O$
Number of students
$11$
$15$
$8$
$6$
From this class, a student is chosen at random. What is the probability that the chosen student has blood group $AB?$
  • A
    $\frac{13}{20}$
  • B
    $\frac{3}{8}$
  • ✓
    $\frac{1}{5}$
  • D
    $\frac{11}{40}$
Answer
Correct option: C.
$\frac{1}{5}$
Total number of students $= 11 + 15 + 8 + 6 = 40$
Number of students having blood group $AB = 8$
$\therefore$ Required probaility $=\frac{8}{40}=\frac{1}{5}$
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MCQ 51 Mark
It is given that the probability of winning a game is $0.7.$ What is the probability of losing the game$?$
  • A
    $0.8$
  • ✓
    $0.3$
  • C
    $0.35$
  • D
    $0.15$
Answer
Correct option: B.
$0.3$
Let $E$ be the event of winning a game.
Then, $P(E) = 0.7$
We know that, $P(E) + P(E') = 1$
Where $E'$ is the event of losing a game.
$⇒ 0.7 + P(E') = 1$
$⇒ P(E') = 0.3$
So, the probability of losing the game is $0.3.$
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MCQ 61 Mark
The table given below shows the month of birth of $36$ students of a class.
Month of birth
Jan.
Feb.
March
April
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
No. of students
$4$
$3$
$5$
$0$
$1$
$6$
$1$
$3$
$4$
$3$
$4$
$3$
A student is chosen at random from the class. What is the probability that the chosen student was born in October$?$
  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{1}{4}$
  • ✓
    $\frac{1}{12}$
Answer
Correct option: D.
$\frac{1}{12}$
Let $E$ be the event of choosing a student born in october.
$P(E) =\frac{\text{Number of students born in october}}{\text{Toatl number of student}}$
$=\frac{3}{36}$
$=\frac{1}{12}$
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MCQ 71 Mark
In a cricket match, a batsman hits a boundary $6$ times out of $30$ balls he plays. What is the probability that in a given delivery, the ball does not hit the boundary$?$
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{5}$
  • ✓
    $\frac{4}{5}$
  • D
    $\frac{3}{4}$
Answer
Correct option: C.
$\frac{4}{5}$
Let $E$ be the event where the ball does not hit a boundary.
The batsman hits a boundary $6$ times.
So, he does not hit a boundary $30 - 6 = 24$ times
$P(E) =\frac{\text{Number of times the ball does not hit a boundary}}{\text{Total number of balls played}}$
$=\frac{24}{30}$
$=\frac{4}{5}$
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MCQ 81 Mark
$80$ bulbs are selected at random from a lot and their lifetime is hours is recorded as under.
Lifetime (in hours)
$300$
$500$
$700$
$900$
$1100$
Frequency
$10$
$12$
$23$
$25$
$10$
One bulb is selected at random from the lot. What is the probability that its life is $1150?$
  • A
    $\frac{1}{80}$
  • B
    $\frac{7}{16}$
  • C
    $1$
  • ✓
    $0$
Answer
Correct option: D.
$0$
Total number of bulbs $= 80$
Number of bulbs having life of $1150$ hours $= 0$
$\therefore$ Required probability $= 0$
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MCQ 91 Mark
In a sample survey of $645$ people, it was found that $516$ people have a high school certificate. If a person is chosen at random, what is the probability that he / she has a high school certificate$?$
  • A
    $\frac{1}{2}$
  • B
    $\frac{3}{5}$
  • C
    $\frac{7}{10}$
  • ✓
    $\frac{4}{5}$
Answer
Correct option: D.
$\frac{4}{5}$
Total number of people $= 645$
Number of people having high school certificate $= 516$
$\therefore$ Required probability $=\frac{516}{645}=\frac{4}{5}$
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MCQ 101 Mark
The outcomes of $65$ throws of a dice were noted as shown below:
Outcome
$1$
$2$
$3$
$4$
$5$
$6$
Number of times
$8$
$10$
$12$
$16$
$9$
$10$
A dice is thrown at random. What is the probability of getting a prime number$?$
  • A
    $\frac{3}{35}$
  • B
    $\frac{3}{5}$
  • ✓
    $\frac{31}{65}$
  • D
    $\frac{36}{65}$
Answer
Correct option: C.
$\frac{31}{65}$
Let $E$ be the event of getting a prime number.
So, $E = \{2, 3, 5\}$
$2$ appears $10$ times, $3$ appears $12$ times and $5$ appears $9$ times.
So, the total number of times a prime number comes up is $10 + 12 + 9 = 31.$
$P(E) =\frac{\text{Number of times a prime number comes up}}{\text{Total number of times the die is thrown}}$
$=\frac{31}{65}$
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MCQ 111 Mark
To know the opinion of the students about the subject Sanskrit, a survey of $200$ students was conducted. The data is recorded as under.
Opinion
like
dislike
Number of students
$135$
$65$
What is the probability that a student chosen at random does not like it$?$
  • A
    $\frac{13}{27}$
  • B
    $\frac{27}{40}$
  • ✓
    $\frac{13}{40}$
  • D
    $\frac{27}{13}$
Answer
Correct option: C.
$\frac{13}{40}$
Total number of students $= 200$
Number of students who does not like Sanskrit $= 65$
$\therefore$ Required probability $=\frac{65}{200}=\frac{13}{40}$
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MCQ 121 Mark
Two coins are tossed $1000$ times and the outcomes are recorded as given below:
Number of heads
$2$
$1$
$0$
Frequency
$200$
$550$
$250$
Now, if two coins are tossed at random, what is the probability of getting at most one head$?$
  • A
    $\frac{3}{4}$
  • ✓
    $\frac{4}{5}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{5}$
Answer
Correct option: B.
$\frac{4}{5}$
Total number of outcomes $= 1000$
Probability of getting at most $1$ head $=$ Probability of getting at most $0$ head $+$ Probability of getting at $1$ head
$=\frac{250}{1000}+\frac{550}{1000}$
$=\frac{800}{1000}$
$=\frac{4}{5}$
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MCQ 131 Mark
A coin is tossed $60$ times and the tail appears 35 times. In a random throw of a coin, what is the probability of getting a head$?$
  • A
    $\frac{7}{12}$
  • B
    $\frac{12}{7}$
  • ✓
    $\frac{5}{12}$
  • D
    $\frac{1}{25}$
Answer
Correct option: C.
$\frac{5}{12}$
Total number of times the coin is tossed $= 60$
Number of times tail appears $= 35$
So, the number of times head appears $= 60 - 35 = 25$
Required probability $=\frac{\text{Number of times head appears}}{\text{Total number of rimes the coin is tossed}}$
$=\frac{25}{60}$
$=\frac{5}{12}$
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MCQ 141 Mark
A bag contains $5$ red, $8$ black and $7$ white balls. One ball is chosen at random. What is the probability that the chosen ball is black?
  • A
    $\frac{2}{3}$
  • ✓
    $\frac{2}{5}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{1}{3}$
Answer
Correct option: B.
$\frac{2}{5}$
Total number of balls in the bag $= 5 + 8 + 7 = 20$
Number of black balls $= 8$
Required probability $=\frac{\text{Number of black balls}}{\text{Total number of balls}}$
$=\frac{8}{20}$
$=\frac{2}{5}$
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MCQ 151 Mark
In $50$ throws of a dice, the outcomes were noted as shown below:
Outcome
$1$
$2$
$3$
$4$
$5$
$6$
Number of times
$8$
$9$
$6$
$7$
$12$
$8$
A dice is thrown at random. What is the probability of getting an even number$?$
  • ✓
    $\frac{12}{25}$
  • B
    $\frac{3}{50}$
  • C
    $\frac{1}{8}$
  • D
    $\frac{1}{2}$
Answer
Correct option: A.
$\frac{12}{25}$
Let $E$ be the event of getting an evan number.
So, $E = \{2, 4, 6\}$
$2$ appears $9$ times, $4$ appears $7$ times and $6$ appears $8$ times.
So, the total number of times an even number comes up is $9 + 7 + 8 = 24$
$P(E) =\frac{\text{Number of times an even number comes up}}{\text{Total number of times the die is thrown}}$
$=\frac{24}{50}$
$=\frac{12}{25}$
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MCQ 161 Mark
A bag contains $16$ cards bearing numbers $1, 2, 3, ..., 16$ respectively. One card is chosen at random. What is the probability that the chosen card bears a number divisible by $3?$
  • A
    $\frac{3}{16}$
  • ✓
    $\frac{5}{16}$
  • C
    $\frac{11}{16}$
  • D
    $\frac{13}{16}$
Answer
Correct option: B.
$\frac{5}{16}$
Total number of cards in the bag $= 16$
Number divisible by $3$ are $\{3, 6, 9, 12, 15\}$
Number of card that bears numbers divisible by $3 = 5$
Required probability $=\frac{5}{16}$
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