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14.Probability
easy
In a certain population $10\%$ of the people are rich, $5\%$ are famous and $3\%$ are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to
A
$0. 07$
B
$0.08$
C
$0. 09$
D
$0. 12$
Solution

(c) Here, $P(R) = \frac{{10}}{{100}} = 0.1$, $P(F) = \frac{5}{{100}} = 0.05$
$P(F \cap R) = \frac{3}{{100}} = 0.03$
$\therefore$ Required probability $= P(R) + P(F) – 2P(F \cap R)$
$= 0.1 + 0.05 -2(0.03) = 0.09$.
Standard 11
Mathematics