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From a pack of 52 playing cards Jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

(i) A black queen

(ii) A red card

(iii) A black jack

(iv) A picture card (Jacks, Queens and Kings are picture cards)

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3 Answers

Total no. of cards = 52

All jacks, queens & kings, aces of red colour are removed.

Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 {remaining cards}

(i) E ⟶ event of getting a black queen

No. of favourable outcomes = 2 {queen of spade & club}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 2/44 = 1/22

(ii) E ⟶ event of getting a red card

No. of favourable outcomes = 26 – 8 = 18 {total red cards – jacks, queens, kings, aces of red colour}

P(E) = 18/44 = 9/22

(iii) E ⟶ event of getting a black jack

No. of favourable outcomes = 2 {jack of club & spade}

P(E) = 2/44 = 1/22

(iv) E ⟶ event of getting a picture card

No. of favourable outcomes = 6 {2 jacks, 2 kings & 2 queens of black colour}

P(E) = 6/44 = 3/22

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We know that,

Total no. of cards = 52

All jacks, queens & kings, aces of red colour are removed.

Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)

(i) Let E = event of getting a black queen

No. of favourable outcomes = 2 (queen of spade & club)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 1/22

(ii)  Let E = event of getting a red card

No. of favourable outcomes = 26 – 8

= 18 (total red cards jacks – queens, kings, aces of red colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 18/44 = 9/22

(iii)  Let E = event of getting a black jack

No. of favourable outcomes = 2 (jack of club & spade)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 2/44 = 1/22

(iv) Let E = event of getting a picture card

No. of favourable outcomes = 6 (2 jacks, 2 kings & 2 queens of black colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 6/44 = 3/22

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Total number of possible outcomes, n(S) = 52 – 2 – 2 – 2 – 2 = 44 

(i) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{44}\) = \(\frac{1}{22}\)

(ii) Number of favorable outcomes, n(E) = 26 – 8 = 18

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{18}{44}\) = \(\frac{9}{22}\)

(iii) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{2}{44}\) = \(\frac{1}{22}\)

(iv) Number of favorable outcomes, 

n(E) = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{6}{44}\) = \(\frac{3}{22}\)

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