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A group of friends wanted to play cards with two identical packs together.
nWhile shuffling the cards, three cards are dropped. Rest of the cards are
nshuffled and one card is drawn at random. Assuming that the dropped
ncards were a queen of hearts, a ten of spades and an ace of clubs, answer
nthe following questions :
n(i) Find the probability that the drawn card is a face card.
n(ii) Find the probability that the drawn card is either a king or a
nqueen.
n(iii) (a) Do you think that the probability of getting a queen was
nhigher if none of the cards were dropped? Justify your
nanswer.
nOR
n(iii) (b) Find the probability that the drawn card is a jack. Compare
nit with the probability when none of the cards were dropped.
nIn which case is the probability of getting a jack higher?
nWhile shuffling the cards, three cards are dropped. Rest of the cards are
nshuffled and one card is drawn at random. Assuming that the dropped
ncards were a queen of hearts, a ten of spades and an ace of clubs, answer
nthe following questions :
n(i) Find the probability that the drawn card is a face card.
n(ii) Find the probability that the drawn card is either a king or a
nqueen.
n(iii) (a) Do you think that the probability of getting a queen was
nhigher if none of the cards were dropped? Justify your
nanswer.
nOR
n(iii) (b) Find the probability that the drawn card is a jack. Compare
nit with the probability when none of the cards were dropped.
nIn which case is the probability of getting a jack higher?
Show SolutionHide Solution↓
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
ndropped.
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 $101 < 104$ (1/2 Mark)
Therefore probability of getting a jack is higher when $3$ cards were
ndropped. (1 Mark)
(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
ndropped.
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 $101 < 104$ (1/2 Mark)
Therefore probability of getting a jack is higher when $3$ cards were
ndropped. (1 Mark)