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Maths 3 Mark Question

Probability · Maharashtra Board STD 8 (MSBSHSE - English Medium). 7 questions.

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3 Mark Question

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Question 13 Marks
In a survey of 200 ladies, it was found that 82 like coffee while 118 dislike it. From these ladies, one is chosen at random. What is the probability that the chosen lady dislikes coffee?
Answer
Total number of ladies = 200
Those who like coffee = 82
Those who dislike coffee = 118
Possible number of outcimes = 200
One lady is chosen at random, then
  1. Probability of a lady who dislikes coffee $=\frac{118}{200}=\frac{59}{100}$
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Question 23 Marks
One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is:
  1. A 4.
  2. A queen.
  3. A black card.
Answer
One card is drawn at random from a deck of well-shuffled deck of 52 card.
Possible outcomes = 52
  1. Probability of a card being a $4=\frac{4}{52}=\frac{1}{13}$
  2. Probability of a card being a queen $=\frac{4}{52}=\frac{1}{13}$
  3. Probability of a card being a black card $=\frac{26}{52}=\frac{1}{2}$
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Question 33 Marks
A box contains 19 balls bearing numbers 1, 2, 3, ..., 19 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is:
  1. A prime number.
  2. An even number.
  3. A number divisible by 3.
Answer
19 ball bearing numbers, 1, 2, 3,.... 19
Possible outcomes = 19
A ball is drawn at random from the box, then
  1. Probability of a ball which bears a prime numbers which are$$ 2, 3, 5, 7, 11, 13 17 and 19 = 8 $=\frac{8}{19}$
  2. Probability of a ball which bears an even number which are$$ 2, 4, 6, 8, 10, 12, 14, 16, 18 = 9 $=\frac{9}{19}$
  3. Probability of a number which bears a number divisible by 3 which are 3, 6, 9, 12, 15, 18 = 6 $=\frac{6}{19}$
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Question 43 Marks
In a single throw of two coins, find the probability of getting:
  1. Both tails,
  2. At least 1 tail,
  3. At the most 1 tail.
Answer
The outcomes when two coins are tossed are HH, HT, TH and TT.
I.e., total no. of possible outcomes = 4
  1. Getting both tails means TT.
Number of outcomes with two tails = 1

$\therefore \text{P}_{\text{both tails}}=\frac{1}{4}$
  1. Getting at least 1 tail means HT, Th and TT.
With at least one tail, total number of outcomes = 3

$\therefore\text{P}_{\text{at least 1 tail}}=\frac{3}{4}$
  1. Getting at most 1 tail means HH, HT and TH.
The number of outcomes for at most 1 tail = 3

$\therefore \text{P}_{\text{at most 1 tail}}=\frac{3}{4}$
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Question 53 Marks
It is known that a box of 100 electric bulbs contains 8 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is:
  1. Defective?
  2. Non-defective?
Answer
In a, box 100 electric bulb, 8 are defective.
Then non-defective bulbs = 100 - 8 = 92
Now possible outcomes = 100
  1. Probability of a drawn bulb, which is defective $=\frac{8}{100}=\frac{2}{25}$
  2. Probability of a drawn bulb which is non defective $=\frac{92}{100}=\frac{23}{25}$
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Question 63 Marks
A bag contains 4 white and 5 blue balls. They are mixed thoroughly and one ball is drawn at random. What is the probability of getting:
  1. A white ball?
  2. A blue ball?
Answer
In a bag, there are 4 ehite and 5 blue balls,
Possible outcomes = 4 +5 = 9
One ball is drawn at random, then
  1. The probability of a white ball $=\frac{4}{9}$
  2. The probability of a blue ball $=\frac{5}{9}$
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Question 73 Marks
A bag contains 5 white, 6 red and 4 green balls. One ball is drawn at random. What is the probability that the ball drawn is:
  1. Green?
  2. White?
  3. Non-red?
Answer
In a bag, there are 5 white, 6 red and 4 green balls.
Possible outcomes is 5 + 6 + 4 = 15
One ball is drawn at random, then
  1. Probability of a green ball $=\frac{4}{15}$
  2. Probability of a white ball $=\frac{5}{15}=\frac{1}{3}$
  3. Probability of a non-red ball $=\frac{5+4}{15}$
$=\frac{9}{15}=\frac{3}{5}$

(5 white and 4 ghreen balls are non-red balls)
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