A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{8}$
$\frac{3}{4}$
Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $
The chances to fail in Physics are $20\%$ and the chances to fail in Mathematics are $10\%$. What are the chances to fail in at least one subject ............ $\%$
From the employees of a company, $5$ persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows :
S.No. | Name | Sex | Age in years |
$1.$ | Harish | $M$ | $30$ |
$2.$ | Rohan | $M$ | $33$ |
$3.$ | Sheetal | $F$ | $46$ |
$4.$ | Alis | $F$ | $28$ |
$5.$ | Salim | $M$ | $41$ |
A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $35$ years?
If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that $2$ white and $1$ black ball will be drawn is
If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$