# All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is

(a) of red color

(b) a queen

(c) an ace

(d) a face card

Answer

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Hint: A standard deck of playing cards has 4 suits of 13 each. And each suit has 3 face cards (a king, a queen, and a jack) so in total 12 cards are present in a standard deck.

Complete step-by-step answer:

As we know a standard deck of playing cards has 52 cards in it broken into 4 suits of 13 cards each.

Each suit has 3 face cards (a king, a queen, and a jack) so, total cards are 12. As, there are two colours present in 52 cards i.e. Red and Black.

So, red cards should be ‘6’ i.e. half of total face cards.

Hence, total cards remained in deck after removing all red face cards will be 46 i.e.,

52-6 = 46

So, total cards present in the deck = 46.

Now, we need to find probability when a card is drawn at random from the remaining cards and the drawn card is:

(a) Of red colour

Total red colour cards in 52 cards are = 26(13 for diamond + 13 for hearts)

Remove cards of red colour = 6

Hence, remaining cards of red colour = 26-6 = 20

As, probability of any event can be given as

$P(E)=\dfrac{\text{Number of favourable cases}}{\text{Total cases}}$

Hence, probability of drawing a red colour card be

$P=\dfrac{20}{46}=\dfrac{10}{23}$

(b) A queen

Total number of queens present in 52 cards = 4(one of each suit).

As queen is a face card, so there will be only 2 queens in the remaining 46 cards as 2 queens (hearts and diamonds) are already removed with the 6 red face cards.

Hence, queens present in remaining cards = 2

So, probability of drawing a queen can be given as

$P=\dfrac{2}{46}=\dfrac{1}{23}$

(c) An ace

Total number of aces in 52 cards = 4 (one for each suit)

As ace is not a face card. So, all the 4 aces will be present in 46 remaining cards after removing all red face cards.

Hence, probability of drawing an ace, we get

$P=\dfrac{4}{46}=\dfrac{2}{23}$

(d) A face card

Total number of face cards present in 52 cards be

= 12 (3 of each suit) or

= 12 (6 of each colour)

Now, when we remove 6 red face cards, then remaining face cards would be 6 each of black colour.

Hence, the total face cards present in 46 cards is 6.

So, probability of drawing face cards from remaining cards will be,

$P=\dfrac{6}{46}=\dfrac{3}{23}$

Note: One can include ‘Ace’ as a face card as well and assume total face cards would be 16 which is wrong. It is a general confusion with students. Hence, face cards include only queen, jack and king and in total 12 cards are there.

All 13 cards include face cards i.e. 1 king, 1 queen, 1 jack and cards numbered from 1,2,3……..10, where 1 is termed as an Ace.

Complete step-by-step answer:

As we know a standard deck of playing cards has 52 cards in it broken into 4 suits of 13 cards each.

Each suit has 3 face cards (a king, a queen, and a jack) so, total cards are 12. As, there are two colours present in 52 cards i.e. Red and Black.

So, red cards should be ‘6’ i.e. half of total face cards.

Hence, total cards remained in deck after removing all red face cards will be 46 i.e.,

52-6 = 46

So, total cards present in the deck = 46.

Now, we need to find probability when a card is drawn at random from the remaining cards and the drawn card is:

(a) Of red colour

Total red colour cards in 52 cards are = 26(13 for diamond + 13 for hearts)

Remove cards of red colour = 6

Hence, remaining cards of red colour = 26-6 = 20

As, probability of any event can be given as

$P(E)=\dfrac{\text{Number of favourable cases}}{\text{Total cases}}$

Hence, probability of drawing a red colour card be

$P=\dfrac{20}{46}=\dfrac{10}{23}$

(b) A queen

Total number of queens present in 52 cards = 4(one of each suit).

As queen is a face card, so there will be only 2 queens in the remaining 46 cards as 2 queens (hearts and diamonds) are already removed with the 6 red face cards.

Hence, queens present in remaining cards = 2

So, probability of drawing a queen can be given as

$P=\dfrac{2}{46}=\dfrac{1}{23}$

(c) An ace

Total number of aces in 52 cards = 4 (one for each suit)

As ace is not a face card. So, all the 4 aces will be present in 46 remaining cards after removing all red face cards.

Hence, probability of drawing an ace, we get

$P=\dfrac{4}{46}=\dfrac{2}{23}$

(d) A face card

Total number of face cards present in 52 cards be

= 12 (3 of each suit) or

= 12 (6 of each colour)

Now, when we remove 6 red face cards, then remaining face cards would be 6 each of black colour.

Hence, the total face cards present in 46 cards is 6.

So, probability of drawing face cards from remaining cards will be,

$P=\dfrac{6}{46}=\dfrac{3}{23}$

Note: One can include ‘Ace’ as a face card as well and assume total face cards would be 16 which is wrong. It is a general confusion with students. Hence, face cards include only queen, jack and king and in total 12 cards are there.

All 13 cards include face cards i.e. 1 king, 1 queen, 1 jack and cards numbered from 1,2,3……..10, where 1 is termed as an Ace.

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