Probability
Probability
Allen Star Batch
Grade 12
Question:
MATCH THE FOLLOWING:
(A) A pack of cards contain 51 cards. Cards are drawn from the pack without replacement. If 1st 13 cards drawn are all red, then the probability that the missing card is black
(B) A pack of cards is counted with face downwards. It is found that one card is missing. One card is drawn and is found to be red. The probability that the missing card is red
(C) A box has 2 white, 4 black and 6 green balls. Person A, draws a ball from it. Then from the remaining balls person B draws two balls which are found to be green. The probability that A has drawn a black ball.
(D) Let p, q be chosen one by one from the set $\{1, \sqrt{2}, \sqrt{3}, 2, e, \pi\}$ with replacement. Now a circle is drawn taking $(p, q)$ as its centre. The probability that at the most two rational points exist on the circle. (Rational points are those points whose both the coordinates are rational).
Step-by-Step Solution
Key Concept: Apply Bayes' theorem with conditional probability to find posterior probabilities. For (A) and (B), use P(missing card color | observed cards) = P(observed | missing color) × P(missing color) / P(observed). For (C), use Bayes' theorem with hypothesis about A's draw and condition on B's two green balls. For (D), involve counting quadratic roots from given set with replacement.
Using Bayes' theorem with $A_1$ (card is red) and $A_2$ (card is black) as the hypothesis, we find $P(\frac{B}{A_1}) = \frac{^{25}C_{13}}{51C_{13}}$ and $P(\frac{B}{A_2}) = \frac{^{26}C_{13}}{51C_{13}}$. Therefore, $P(\frac{A_2}{B}) = \frac{P(A_2) \cdot P(\frac{B}{A_2})}{P(A_1) \cdot P(\frac{B}{A_1}) + P(A_2) \cdot P(\frac{B}{A_2})} = \frac{^{26}C_{13}}{^{25}C_{13} + ^{26}C_{13}}$.
Correct Answer: [A-t][B-q] [C-s][D-r]