Fill In The Blanks[1 Marks ]

Question 11 Mark

Answer

If A and B are two events associated with a random experiment such that P(A) = 0.3, P(B) = 0.2 and $\text{P}(\text{A}\cap\text{B})=0.1,$ then the value of $\text{P}(\text{A}\cap\bar{\text{B}})$ is 0.2
Solution:
$\text{P}(\text{A}\cap\bar{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$=0.3-0.1=0.2$

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Question 21 Mark

Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then $\bar{\text{E}}$is __________.

Answer

Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then $\bar{\text{E}}$is {2, 4, 6}.
Solution:
Here, S = {1, 2, 3, 4, 5, 6}
and E = {1, 3, 5}
$\therefore\ \bar{\text{E}}=\text{S}-\text{E}=\{2,4,6\}$

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Question 31 Mark

If $e_1, e_2, e_3, e_4$ are the four elementary outcomes in a sample space and $P(e_1) = 0.1, P(e_2) = 0.5, P (e_3) = 0.1,$ then the probability of $e_4$ is ___________.

Answer

If $e_1, e_2, e_3, e_4 $are the four elementary outcomes in a sample space and $P(e_1) = 0.1, P(e_2) = 0.5, P (e_3) = 0.1,$ then the probability of $e_4$ is $0.3$
$\text{P}(\text{E}_1)+\text{P}(\text{E}_2)+\text{P}(\text{E}_3)+\text{P}(\text{E}_4)=1$
$\Rightarrow0.1+0.5+0.1+\text{P}(\text{E}_4)=1$
$\Rightarrow0.7+\text{P}(\text{E}_4)=1$
$\therefore\ \text{P}(\text{E}_4)=0.3$

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Question 41 Mark

The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is __________.

Answer

The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is 0.2
Solution:
$\text{P}(\bar{\text{A}}\cap\bar{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-\big[\text{P(A)}+\text{P(B)}\big]$ [Since, A and B are mutually exclusive]
$=1-(0.5+0.3)$
$=1-0.8=0.2$

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Question 51 Mark

The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is __________.

Answer

The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is 0.15
Solution:
P(losing) = 1 - (0.77 + 0.08) = 0.15

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Maths Fill In The Blanks[1 Marks ]

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