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Maths 2 Mark Question

Data Handling - IV (Probability) · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 21 questions.

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Three coins are tossed together. Find the probability of getting:
no tails.
Answer
When 3 coins are tossed together, the outcomes are as follows:
S = {(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8.
Let A be the event of getting triplets having no tail.
Triplets having no tail: (h, h, h)Therefore, the total number of favourable outcomes is 1.
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{1}{8}$
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Question 22 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a doublet of odd numbers.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting doublets of odd numbers in the sample space.
The doublets of odd numbers in the sample space are (1, 1), (3, 3) and (5, 5).
Hence, the number of favourable outcomes is 3.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{36}=\frac{1}{12}$
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Question 32 Marks
What is the probability that a number selected from the numbers 1, 2, 3, _____, 15 is a multiple of 4?
Answer
There are 15 numbers from 1, 2, ______, 15.
Hence, the total number of cases is 15.
Again, the multiples of 4 are 4, 8 and 12.
Therefore, the total number of favourable cases is 3.
$\therefore$ (the number is a multiple of 4)
$=\frac{\text{Number of favourable casesTotal}}{\text{ number of cases} }=\frac{3}{15}=\frac{1}{5}$
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Question 42 Marks
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Answer
Number of red balls = 10
Number of white balls = 8
Total number of balls in the urn = 10 + 8 = 18
Therefore, the total number of cases is 18 and the number of favourable cases is 8.
$\therefore\text{P}\text{ (The ball drawn is white) }=\frac{\text{Number of favourable cases}}{\text{Total number of cases}}\\=\frac{8}{18}=\frac{4}{9}$
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Question 52 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a sum less than 7.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is less than 7.
The pairs whose sum is less than 7 are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) and (5, 1).
Hence, the number of favourable outcomes is 15.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}\\$
$=\frac{15}{36}=\frac{5}{12}$
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Question 62 Marks
Three coins are tossed together. Find the probability of getting:
at least two heads.
Answer
When 3 coins are tossed together, the outcomes are as follows:
S = {(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8.
Let A be the event of getting triplets having at least 2 heads.
Triplets having at least 2 heads: (h, h, t), (h, t, h), (t, h, h), (h, h, h)Therefore, the total number of favourable outcomes is 4.
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$$=\frac{4}{8}=\frac{1}{2}$
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Question 72 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a sum greater than 9.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is greater than 9.
The pairs whose sum is greater than 9 are (4, 6), (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6).
Hence, the number of favorable outcomes is 6.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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Question 82 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a sum less than 6.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is less than 6.
The pairs whose sum is less than 6 are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) and (4, 1).
Hence, the number of favourable outcomes is 10.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{10}{36}=\frac{5}{18}$
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Question 92 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a sum more than 7.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is more than 7.
The pairs whose sum is more than 7 are (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6).
Hence, the number of favourable outcomes is 15.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}\\$$=\frac{15}{36}=\frac{5}{12}$
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Question 102 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a doublet of prime numbers.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting doublets of prime numbers in the sample space.
The doublets of prime numbers in the sample space are (2, 2), (3, 3) and (5, 5).
Hence, the number of favourable outcomes is 3.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{36}=\frac{1}{12}$
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Question 112 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
an even number on first.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs who has even numbers on first in the sample space.
The pairs who has even numbers on first are: (2, 1), (2, 2), _____ (2, 6), (4, 1), _____ ,(4, 6), (6, 1), _____ (6, 6).
Hence, the number of favourable outcomes is 18.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{18}{36}=\frac{1}{6}$
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Question 122 Marks
Three coins are tossed together. Find the probability of getting:
at least one head and one tail.
Answer
When 3 coins are tossed together, the outcomes are as follows:
S = {(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8.
Let A be the event of getting triplets having at least one head and one tail.Triplets having at least one head and one tail: (h, h, t), (h, t, h), (t, h, h), (h, h, t), (t, t, h), (t, h, t) Therefore, the total number of favourable outcomes is 6.
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$$=\frac{6}{8}=\frac{3}{4}$
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Question 132 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a number other than 5 on any dice.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs that has the number 5
The pairs that has the number 5 are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4) and (6, 6).
Hence, the number of favourable outcomes is 11.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{11}{36}$
$\therefore\text{P}({\bar{\text{A}}})=1−\text{P}\text{(A)}=1−\frac{11}{36}=\frac{25}{36}​​​​​​​$
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Question 142 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
neither 9 nor 11 as the sum of the numbers on the faces.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is 9 or 11.
The pairs whose sum is 9 are (3, 6), (4, 5), (5, 4) and (6, 3).And, the pairs whose sum is 11 are (5, 6) and (6, 5).
Hence, the number of favourable outcomes is 6.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
$\therefore$ P(sum of the pairs with neither 9 nor 11) = 1 - P(sum of the pairs having 9 or 1)
$=1-\frac{1}{6}=\frac{5}{6}$
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Question 152 Marks
The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
Answer
Let A be the event of raining tomorrow.
The probability that it will rain tomorrow, P(A), is 0.85.
Since the event of raining tomorrow and not raining tomorrow are complementary to each other, the probability of not raining.
$\text{P}(\bar{\text{A}}) = 1 − \text{P(A)} = 1 − 0.85 = 0.15$
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Question 162 Marks
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?
Answer
Number of red balls = 6
Number of black balls = 8
Numberof white balls = 4
Total number of balls = 6 + 8 + 4 = 18
$\therefore$ Total number of cases = 18
Again, number of balls that are not black = 18 - 8 = 10
Thus, the number of favourable cases is 10.
$\therefore$ (the drawn ball is not black)
$=\frac{\text{Number of favourable casesTotal}}{\text{number cases}}=\frac{10}{18}=\frac{5}{9}$
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Question 182 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
a doublet.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting doublets in the sample space.
The doublets in the sample space are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
Hence, the number of favourable outcomes is 6.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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Question 192 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:
8 as the sum.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1,4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs whose sum is 8.Now, the pairs whose
sum is 8 are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
Therefore, the total number of favourable outcomes is 5.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{5}{36}$
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Question 202 Marks
Three coins are tossed together. Find the probability of getting:
exactly two heads.
Answer
When 3 coins are tossed together, the outcomes are as follows:
S = {(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8.
Let A be the event of getting triplets having exactly 2 heads.
Triplets having exactly 2 heads: (h, h, t), (h, t, h), (t, h, h)
Therefore, the total number of favourable outcomes is 3.
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{8}$
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Question 212 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:an even number on one and a multiple of 3 on the other.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), ______ (6, 5), (6, 6)}
Hence, the total number of outcomes is 36.
Let A be the event of getting pairs with an even number on one die and a multiple of 3 on the other.
The pairs with an even number on one die and a multiple of 3 on the other are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6).
Hence, the number of favourable outcomes is 6.
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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