AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(B) Let $E_1$ be the event that $A$ speaks the truth.
$P(E_1) = \frac{20}{100} = \frac{1}{5}$.
Then,$P(\overline{E_1}) = 1 - \frac{1}{5} = \frac{4}{5}$.
Let $E_2$ be the event that $B$ speaks the truth.
$P(E_2) = \frac{80}{100} = \frac{4}{5}$.
Then,$P(\overline{E_2}) = 1 - \frac{4}{5} = \frac{1}{5}$.
Their statements do not match if ($A$ speaks the truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks the truth).
Required Probability $= P(E_1) \cdot P(\overline{E_2}) + P(\overline{E_1}) \cdot P(E_2)$.
$= (\frac{1}{5} \times \frac{1}{5}) + (\frac{4}{5} \times \frac{4}{5})$.
$= \frac{1}{25} + \frac{16}{25} = \frac{17}{25}$.

(D) Total number of bulbs = $50$.
Number of defective bulbs = $5$.
Number of non-defective bulbs = $50 - 5 = 45$.
We need to draw $3$ bulbs without replacement.
The probability that the first bulb is non-defective is $P(E_1) = \frac{45}{50}$.
After drawing one non-defective bulb,there are $44$ non-defective bulbs left out of $49$ total bulbs.
The probability that the second bulb is non-defective is $P(E_2|E_1) = \frac{44}{49}$.
After drawing two non-defective bulbs,there are $43$ non-defective bulbs left out of $48$ total bulbs.
The probability that the third bulb is non-defective is $P(E_3|E_1 \cap E_2) = \frac{43}{48}$.
The probability that all $3$ bulbs are non-defective is $P = \frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}$.
$P = \frac{9}{10} \times \frac{44}{49} \times \frac{43}{48} = \frac{9 \times 11 \times 43}{10 \times 49 \times 12} = \frac{3 \times 11 \times 43}{10 \times 49 \times 4} = \frac{1419}{1960}$.