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 $P\left(E^{\prime} \cap F^{\prime}\right)$ $=P(E \cup F)^{\prime}$ $=1-P(E \cup F)=1-0.13=0.87$
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A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
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 $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $
If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are independent, then $x =$