Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both balls are red.
Total number of balls $=18$
Number of red balls $=8$
Number of black balls $=10$
Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
$\therefore$ Probability of getting a red ball in the second draw $=\frac{8}{18}=\frac{4}{9}$
Therefore, probability of getting both the balls red $=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}$
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10 .$ The probability that both will qualify the examination is $0.02 .$ Find the probability that Atleast one of them will not qualify the examination.
Consider three sets $E_1=\{1,2,3\}, F_1=\{1,3,4\}$ and $G_1=\{2,3,4,5\}$. Two elements are chosen at random, without replacement, from the set $E _1$, and let $S _1$ denote the set of these chosen elements.
Let $E_2=E_1-S_1$ and $F_2=F_1 \cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.
Let $G _2= G _1 \cup S _2$. Finally, two elements are chosen at random, without replacement, from the set $G _2$ and let $S _3$ denote the set of these chosen elements.
Let $E_3=E_2 \cup S_3$. Given that $E_1=E_3$, let $p$ be the conditional probability of the event $S_1=\{1,2\}$. Then the value of $p$ is
Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are
If $P(A) = 0.25,\,\,P(B) = 0.50$ and $P(A \cap B) = 0.14,$ then $P(A \cap \bar B)$ is equal to
Let $A$ and $B$ be two events such that $P\overline {(A \cup B)} = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P(\bar A) = \frac{1}{4},$ where $\bar A$ stands for complement of event $A$. Then events $A$ and $B$ are