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Question

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)

(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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Solution

Total no. of cards in a pack =52

After removing red-colored - jack, queen, king, and aces,

No. of cards =52−8=44 and

No. of red cards =26−8=18.

**Solution(i):**

**Solution(ii):**

**Solution(iii):**

No. of black queens =2

Therefore, 2C1( Selecting 1 out of 2 items) times out of 44C1( Selecting 1 out of 44 items) a black queen is picked.

Let E be the event of getting a black queen from pack

We know that, Probability P(E) =(No.of favorable outcomes)(Total no.of possible outcomes)=2C144C1=244=122

No. of red cards =18

Therefore, 18C1( Selecting 1 out of 18 items) times out of 44C1( Selecting 1 out of 44 items) a red card is picked.

Let E be the event of getting a red card from the pack

We know that, Probability P(E) =(No.of favorable outcomes)(Total no.of possible outcomes)=18C144C1=1844=922

No. of black jack cards =2

**Solution(iv):**

Therefore, 2C1( Selecting 1 out of 2 items) times out of 44C1( Selecting 1 out of 44 items) a black jack card is picked.

Let E be the event of getting a black jack card from the pack

We know that, Probability P(E) =(No.of favorable outcomes)(Total no.of possible outcomes)=2C144C1=244=122

No. of picture cards =12−6=6 ..... (As red cards are removed and there are 6 red picture cards )

Therefore, 6C1( Selecting 1 out of 6 items) times out of 44C1( Selecting 1 out of 44 items) a picture card is picked.

Let E be the event of getting a picture card from pack

We know that, Probability P(E) =(No.of favorable outcomes)(Total no.of possible outcomes)=6C144C1=644=322

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