Three students $A$,$B$ and $C$ are running a race. $A$ and $B$ have the same probability of winning and each is twice as likely to win as $C$. Then,the probability that $B$ or $C$ wins is equal to (assuming there are no ties)
- A$\frac{2}{5}$
- B$\frac{3}{5}$
- C$\frac{3}{7}$
- D$\frac{2}{7}$
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Similar Questions
Let $\Omega$ be the sample space and $A \subseteq \Omega$ be an event. Given below are two statements:
$(S1) : \text{If } P(A) = 0, \text{ then } A = \phi$
$(S2) : \text{If } P(A) = 1, \text{ then } A = \Omega$
Then:
$(S1) : \text{If } P(A) = 0, \text{ then } A = \phi$
$(S2) : \text{If } P(A) = 1, \text{ then } A = \Omega$
Then:
DifficultJEE MAIN 2023
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. The correct match of List $I$ from List $II$ is:
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 $II$ |
| $(i) \ P(A \cap B)$ | $(1) \ 0.4$ |
| $(ii) \ P(A \cap \bar{B})$ | $(2) \ 0.2$ |
| $(iii) \ P(\bar{A} \cap B)$ | $(3) \ 0.3$ |
| $(iv) \ P(\bar{A} \cap \bar{B})$ | $(4) \ 0.1$ |
Three randomly chosen non-negative integers $x, y$ and $z$ are found to satisfy the equation $x+y+z=10$. Then the probability that $z$ is even,is
Three coins are tossed once. Find the probability of getting at most two tails.
Easy
View Solution$A$ letter is chosen at random from the word $ASSASSINATION$. Find the probability that the letter is a vowel. (in $/13$)
Easy
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