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Question 11 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

0.2, because
Given, $P(A)=0.5, P(B)=0.3$
and $P(A \cap B)=0$
$[\because A$ and $B$ are mutually exclusive events]
Now, Probability that neither $A$ nor $B=P(\bar{A} \cap \bar{B})$
$\begin{array}{l}=P(\overline{A \cup B}) \\ =1-[P(A \cup B)] \\ =1-[P(A)+P(B)-P(A \cap B)] \\ =1-[0.5+0.3-0] \\ =1-0.8=0.2\end{array}$

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Question 21 Mark

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

Answer

0.2 , because
Given, $P(A)=0.3, P(B)=0.2$ and $P(A \cap B)=0.1$
We know that,
$P(A \cap \bar{B})=P(A)-P(A \cap B)$
$\Rightarrow \quad P(A \cap \bar{B})=0.3-0.1=0.2$

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Question 31 Mark

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

Answer

0.3 , because
We know that the sum of all probabilities $=1$
$
\therefore P\left(e_1\right)+P\left(e_2\right)+P\left(e_3\right)+P\left(e_4\right)=1
$
$\Rightarrow 0.1+0.5+0.1+P\left(e_4\right)=1$
$\Rightarrow P\left(e_4\right)=1-(0.7)$
$\Rightarrow P\left(e_4\right)=0.3$

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Question 41 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

0.15 , because
Given, $P$ (loosing the game) $=1-[P$ (winning football game $)+P($ the game will tie $)]$
$\begin{array}{l}=1-(0.77+0.08) \\ =1-0.85 \\ =0.15\end{array}$

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