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 Both Anil and Ashima will not qualify the examination.
Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination, respectively. Given that
$P(E)=0.05$, $P(F)=0.10$ and $P(E \cap F)=0.02$
Then
The event ' both Anil and Ashima will not qualify the examination' may be expressed as $E ^{\prime} \cap F^{\prime}$
since, $E ^{\prime}$ is 'not $E^{\prime},$ i.e., Anil will not qualify the examination and $F ^{\prime}$ is 'not $F^{\prime}$, i.e. Ashima will not qualify the examination.
Also $E ^{\prime} \cap F ^{\prime}=( E \cup F )^{\prime}$ (by Demorgan's Law)
Now $P ( E \cup F )= P ( E )+ P ( F )- P ( E \cap F )$
or $P(E \cup F)=0.05+0.10-0.02=0.13$
Therefore $P\left(E^{\prime} \cap F^{\prime}\right)$ $=P(E \cup F)^{\prime}$ $=1-P(E \cup F)=1-0.13=0.87$
Let $A$ and $B$ be independent events such that $\mathrm{P}(\mathrm{A})=\mathrm{p}, \mathrm{P}(\mathrm{B})=2 \mathrm{p} .$ The largest value of $\mathrm{p}$, for which $\mathrm{P}$ (exactly one of $\mathrm{A}, \mathrm{B}$ occurs $)=\frac{5}{9}$, is :
Let $A$,$B$ and $C$ be three events such that $P\left( {A \cap \bar B \cap \bar C} \right) = 0.6$, $P\left( A \right) = 0.8$ and $P\left( {\bar A \cap B \cap C} \right) = 0.1$, then the value of $P$(atleast two among $A$,$B$ and $C$ ) equals
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$\frac {1}{3}$ | $\frac {1}{5}$ | $\frac {1}{15}$ | ........ |
A card is drawn from a pack of cards. Find the probability that the card will be a queen or a heart
The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is