$A$ bag contains $5$ black balls, $4$ white balls and $3$ red balls. If a ball is selected at random, the probability that it is a black or a red ball, is
- A$\frac{1}{3}$
- B$\frac{1}{4}$
- C$\frac{5}{12}$
- D$\frac{2}{3}$
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Similar Questions
$A$ six-faced unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed to be $7$. The probability that the number $3$ has appeared at least once is:
DifficultTS EAMCET 2014
View SolutionWhen a die is thrown,let event $A$ be getting a number greater than $3$,and let event $B$ be getting a number less than $5$. What is $P(A \cup B)$?
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View SolutionWhat is the probability that a randomly selected leap year has $53$ Sundays or $53$ Mondays? Express your answer as the numerator of the fraction (in $/7$).
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View Solution$A$ card is drawn at random from a pack of $100$ cards numbered $1$ to $100$. The probability of drawing a number which is a perfect square is
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View SolutionLet $A$ and $B$ be events in a sample space $S$ such that $P(A)=0.5$, $P(B)=0.4$ and $P(A \cup B)=0.6$. Observe the following lists. Match List-$I$ with List-$II$ and select the correct option.
List-$I$ List-$II$ $(i) \ P(A \cap B)$ $(1) \ 0.4$ $(ii) \ P(A \cap \bar{B})$ $(2) \ 0.2$ $(iii) \ P(\bar{A} \cap B)$ $(3) \ 0.3$ $(iv) \ P(\bar{A} \cap \bar{B})$ $(4) \ 0.1$
| List-$I$ | List-$II$ |
| $(i) \ P(A \cap B)$ | $(1) \ 0.4$ |
| $(ii) \ P(A \cap \bar{B})$ | $(2) \ 0.2$ |
| $(iii) \ P(\bar{A} \cap B)$ | $(3) \ 0.3$ |
| $(iv) \ P(\bar{A} \cap \bar{B})$ | $(4) \ 0.1$ |
DifficultAP EAMCET 2011
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