CBSE 12th Standard Maths Subject Probability Ncert Exemplar 4 Marks Questions With Solution 2021
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CBSE 12th Standard Maths Subject Probability Ncert Exemplar 4 Marks Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Maths
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A committee of 4 students is selected at random from a group consisting of 8 boys and 4 girls. Given that there is at least one girl in the committee, calculate the probability that there are exactly 2 girls in the committee.
(a) -
A and B throw a pair of die alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. If A starts the game, find the probability of winning the game by A in third throw of pair of dice.
(a) -
Bag 1 contains 3 black and 2 white balls, bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
(a) -
A biased die is such that \(P(4)=\frac{1}{10}\) and other scores-being equally likely. The die is tossed twice. If X is the 'number of four seen', then find the variance of the random variable X.
(a) -
Suppose 10000 tickets are sold in a lottery each for Rs. 1. First prize is of Rs. 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, then what is your expectation?
(a)
4 Marks
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CBSE 12th Standard Maths Subject Probability Ncert Exemplar 4 Marks Questions With Solution 2021 Answer Keys
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Let the events be As:
A : At least one girl is chosen
B : At least 2 girls are chosen
We requireP(B/A)
\(P(\overset { - }{ A } )\)= P(No girl is chosen)
\(=\frac { ^{ 8 }{ C }_{ 4 } }{ ^{ 12 }{ C }_{ 4 } } =\frac { 70 }{ 495 } =\frac { 14 }{ 99 } \)
\(P(A)=1-P(\overset { - }{ A } )=1-\frac { 14 }{ 99 } =\frac { 85 }{ 99 } \)
And \(P(A\cap B)\)= P(2 boys and 2 girls) = \(\frac { ^{ 8 }{ C }_{ 2 }.^{ 4 }{ C }_{ 2 } }{ ^{ 12 }{ C }_{ 4 } } \)
\(=\frac { 6\times 28 }{ 495 } =\frac { 56 }{ 165 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 56 }{ 165 } \times \frac { 99 }{ 85 } =\frac { 168 }{ 425 } \) -
Here \(P(A)=\frac { 5 }{ 36 } \)
[(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)]
\(P(\overset { - }{ A } )=1-\frac { 5 }{ 36 } =\frac { 31 }{ 36 } \)
And \(P(B)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
[(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)]
\(P(\overset { - }{ B) } =1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)
Reqd.probability = \(P(\overset { \_ }{ A } \overset { \_ }{ B } A)\)
\(=P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )P(A)\)
\(=\frac { 31 }{ 36 } \times \frac { 5 }{ 6 } \times \frac { 5 }{ 36 } =\frac { 775 }{ 7776 } \) -
Let EI = Bag I is selected
E 2 = Bag II is selected
and A = Black ball is drawn
Then,\(P\left(E_{1}\right)=P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{A}{E_{1}}\right)=\frac{3}{5}, P\left(\frac{A}{E_{2}}\right)=\frac{2}{6}\)
\(\therefore\) Required probability
\(P(A)=P\left(E_{1}\right) \times P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \times P\left(\frac{A}{E_{2}}\right)\)
= \(\frac{7}{15}\) -
Since, X = Number of four seen
On tossing two dice, X = 0, 1, 2
Also, \(P(4)=\frac{1}{10} \text { and } P(\operatorname{not} 4)=\frac{9}{10}\)
So, \(P(X=0)=P(\operatorname{not} 4) \cdot P(\operatorname{not} 4)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}\)
\(P(X=1)=P(\operatorname{not} 4) \cdot P(4)+P(4) \cdot P(\operatorname{not} 4)\)
\(=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100}\)
\(P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}\)
Thus, we get the following table
\(\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & 81 / 100 & 18 / 100 & 1 / 100 \\ \hline \end{array}\)
\(\begin{array}{c|c|c|c} \hline X P(X) & 0 & 18 / 100 & 2 / 100 \\ \hline X^{2} P(X) & 0 & 18 / 100 & 4 / 100 \\ \hline \end{array}\)
\(\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}\)
\(=\left(0+\frac{18}{100}+\frac{4}{100}\right)-\left(0+\frac{18}{100}+\frac{2}{100}\right)^{2}\)
\(=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\)
\(=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18\) -
\(\begin{array}{c|l|l|l|l} \hline x & 0 & 500 & 2000 & 3000 \\ \hline P(X) & \frac{9995}{10000} & \frac{3}{10000} & \frac{1}{10000} & \frac{1}{10000} \\ \hline P_{i} X_{1} & 0 & \frac{1500}{10000} & \frac{2000}{10000} & \frac{3000}{10000} \\ \hline \end{array}\)
\(\because E(X)=\Sigma X P(X)\)
= 0.65
4 Marks
