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A group of friends wanted to play cards with two identical packs together. While shuffling the cards, three cards are dropped. Rest of the cards are shuffled and one card is drawn at random. Assuming that the dropped cards were a queen of hearts, a ten of spades and an ace of clubs, answer the following questions:
(i) Find the probability that the drawn card is a face card.
(ii) Find the probability that the drawn card is either a king or a queen.
(iii) (a) Do you think that the probability of getting a queen was higher if none of the cards were dropped? Justify your answer.
OR
(iii) (b) Find the probability that the drawn card is a jack. Compare it with the probability when none of the cards were dropped. In which case is the probability of getting a jack higher?
(i) Find the probability that the drawn card is a face card.
(ii) Find the probability that the drawn card is either a king or a queen.
(iii) (a) Do you think that the probability of getting a queen was higher if none of the cards were dropped? Justify your answer.
OR
(iii) (b) Find the probability that the drawn card is a jack. Compare it with the probability when none of the cards were dropped. In which case is the probability of getting a jack higher?
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Sol. Total number of cards $= 2 \times 52 - 3 = 101$
(i) P (a face card) $= \frac{23}{101}$ (1 Mark)
(ii) P (either a king or a queen) $= \frac{15}{101}$ (1 Mark)
(iii) (a)Yes (1/2 Mark)
P (a queen when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
P (a queen when cards were dropped) $= \frac{7}{101}$ (1/2 Mark)
$\therefore \frac{8}{104} > \frac{7}{101}$ as $808 > 728$ (1/2 Mark)
So probability of getting a queen was higher if none of the cards were dropped.
OR
(iii) (b) P (a jack when cards were dropped) $= \frac{8}{101}$ (1/2 Mark)
P (a jack when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
Since $\frac{8}{101} > \frac{8}{104}$ as $101 < 104$ (1 Mark)
Therefore probability of getting a jack is higher when $3$ cards were dropped.
(i) P (a face card) $= \frac{23}{101}$ (1 Mark)
(ii) P (either a king or a queen) $= \frac{15}{101}$ (1 Mark)
(iii) (a)Yes (1/2 Mark)
P (a queen when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
P (a queen when cards were dropped) $= \frac{7}{101}$ (1/2 Mark)
$\therefore \frac{8}{104} > \frac{7}{101}$ as $808 > 728$ (1/2 Mark)
So probability of getting a queen was higher if none of the cards were dropped.
OR
(iii) (b) P (a jack when cards were dropped) $= \frac{8}{101}$ (1/2 Mark)
P (a jack when no cards were dropped) $= \frac{8}{104}$ (1/2 Mark)
Since $\frac{8}{101} > \frac{8}{104}$ as $101 < 104$ (1 Mark)
Therefore probability of getting a jack is higher when $3$ cards were dropped.