Two students Anil and Ashima appeared in an examination. probability that Anil will qualify examinat

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14.Probability

medium

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 Only one of them will qualify the examination.

$0.11$

Solution

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 only one of them will qualify the examination is same as the event either (Anil will qualify, andAshima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., $E \cap F ^{\prime}$ or $E ^{\prime} \cap F ,$ where $E \cap F ^{\prime}$ and $E ^{\prime} \cap F$ are mutually exclusive.

Therefore, $P$ (only one of them will qualify) $=P(E \cap F^{\prime} $ or $E^{\prime} \cap F)$

$= P \left( E \cap F ^{\prime}\right)$ $+ P \left( E ^{\prime} \cap F \right)$ $= P ( E )- P ( E \cap F )+ P ( F )- P ( E \cap F ) $

$=0.05-0.02+0.10-0.02=0.11$

Standard 11

Mathematics

Fill in the blanks in following table :

$P(A)$ $P(B)$ $P(A \cap B)$ $P (A \cup B)$

$0.35$ ……….. $0.25$ $0.6$

easy

Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is

normal

(KVPY-2014)

In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random. If she reads English newspaper, find the probability that she reads Hindi newspaper.

medium

Let $S$ be a set containing n elements and we select $2$ subsets $A$ and $B$ of $S$ at random then the probability that $A \cup B = S$ and $A \cap B = \phi $ is

hard

$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.