CBSE 10th Standard Maths Subject Probability HOT Questions 2 Mark Questions With Solution 2021
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CBSE 10th Standard Maths Subject Probability HOT Questions 2 Mark Questions With Solution 2021
10th Standard CBSE

Reg.No. :
Maths

A bag contains 12 marbles out of which y are white.
(i)If one marble is drawn at random from the bag, what is the probability that it will be white marble?
(ii)If 6 more white marbles are put in the bag, the probability of drawing a white marble will double than in part (i), find y.(a) 
Two dice are thrown at the same time.Find the probability of getting different numbers on the dice.
(a) 
From a group of 3 Girls and 2 Boys, two children are selected at random.Find the probability such that at least one boy is selected.
(a) 
In a bagA, there are four cards numbered 1,3,5 and 7 respectively.In another bagB, there are three cards numbered 2, 4 and 6 respectively.A card is drawn at random from each bag.
(i)Write the possible outcomes i.e., sample space
(ii)Find the probability that the sum of these two cards drawn is:
(a)7
(b)even
(c)odd
(d)more than 7(a)
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CBSE 10th Standard Maths Subject Probability HOT Questions 2 Mark Questions With Solution 2021 Answer Keys

Here, total number of marbles = 12
Number of white marbles = y
(i) P(a white marble) = \(\frac { y }{ 12 } \)
(ii) Now, 6 more white marbles are added in the bag.
\(\therefore\)Total number Of marbles = 12 + 6 = 18
Number of white marbles = y + 6 .
\(\therefore\) P(a white marble) = \(\frac { y+6 }{ 18 } \)
According to the statement of the question, we have
\(\frac { y+6 }{ 18 } =2(\frac { y }{ 12 } )\)
\(\Rightarrow \frac { y+6 }{ 3 } =y\)
\(\Rightarrow \) 3y=y+6
\(\Rightarrow \) 2y=6
\(\Rightarrow \) y=3 
Since the two dice are thrown simultarrously Total number Of outconrs = 6 x 6 = 36
Number of outconrs for getting same numbers on both dice = 6
\(\Rightarrow\)P(getting same number) = \(\frac { 6 }{ 36 } = \frac { 1 }{ 6 } \)
Now, P(getting same numbers) + P(getting different numbers) = 1
\(\Rightarrow \frac{ 1 } {6}\) + P(getting different numbers)= 1
\(\Rightarrow\)P(gerting different numbers) = 1\(\frac {1} {6} = \frac {5} {6}\) 
Let G_{1}, G_{2}, G_{3} and B_{1}, B_{2} be three girls and two boys respectively.
Since two children are selected at random, therefore the following are the possible groups:
B_{1}B_{2}, B_{1}G_{1}, B_{1}G_{2}, B_{1}G_{3}, B_{2}G_{1}, B_{2}G_{2}, B_{2}G_{3}, G_{1}G_{2}, G_{1}G_{3}, G_{2}G_{3}
Total number of cases = 10
Now, at least one boy is selected in the following ways:
B_{1}B_{2}, B_{1}G_{1}, B_{1}G_{2}, B_{1}G_{3}, B_{2}G_{1}, B_{2}G_{2}, B_{2}G_{3}
Total number of favourable cases = 7
\(\therefore\) Required probability = \(\frac {7} {10}\) 
There are 12 (4 x 3) possible outcomes as listed below
(1,2), (1,4), (1,6), (3, 2), (3,4), (3, 6), (5, 2), (5,4), (5,6), (7,2). (7, 4) (7, 6) (ii) (a) only (1, 6), (3,.4), (5, 2) given sum as 7
\(\therefore\)Required probability = \(\frac {2} {13}=\frac {1} {4}\)
(b) There is no even sum
\(\therefore\)Required probability= \(\frac {0} {12}=0\)
(c) All the sums are odd
\(\therefore\)Required probability =\(\frac {12} {12} =1\)
(d) only (3, 6), (5, 4), (5, 6), (7, 2). (7, 4) and (7, 6) gives sum as more than 7
\(\therefore\)Required probability = \(\frac {6} {12}\)
=\(\frac {1} {2}\)