Questions & Answers

Question

Answers

$\left( i \right)$ A black queen $\left( {ii} \right)$ A red card $\left( {iii} \right)$ A black jack $\left( {iv} \right)$ A picture card

(Jacks, kings and queen are picture cards)

Answer

Verified

156.6k+ views

Hint: In this question first removed from a pack of 52 playing cards jacks, queens, kings and aces of red color, then calculated the remaining cards which is the total outcomes, then find out the favorable outcomes for all different cases, then apply the formula of probability, so use these concepts to reach the solution of the question.

Complete step-by-step answer:

In a deck of 52 cards we have 2 red queens, 2 red jacks, 2 red kings, 2 red aces one of hearts and one of diamond and we have removed them both so the total cards removed are

$\left( {2 + 2 + 2 + 2} \right) = 8$

So, the remaining cards are $\left( {52 - 8} \right) = 44$.

Therefore total outcomes $ = 44$

$\left( i \right)$ A black queen

In the remaining cards there are only 2 black queens, one of spade and one of club.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{2}{{44}} = \dfrac{1}{{22}}$.

$\left( {ii} \right)$ A red card

As we know that in a pack of 52 cards there are 4 houses having 13 cards each. (Spade, heart, diamond and club)

Therefore the total number of red cards is $\left( {13 + 13} \right) = 26$ (heart and diamond).

But we have removed 8 red cards, so the total number of red cards remaining are $\left( {26 - 8 = 18} \right)$.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{{18}}{{44}} = \dfrac{9}{{22}}$.

$\left( {iii} \right)$ A black jack

In the remaining cards there are only 2 black jacks, one of spade and one of club.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{2}{{44}} = \dfrac{1}{{22}}$.

$\left( {iv} \right)$ A picture card

As we know in a pack of 52 cards (Jacks, kings and queens are picture cards), so, there are 4 jacks, 4 kings and 4 queens in a pack of 52 cards.

So the total picture card becomes $\left( {4 + 4 + 4 = 12} \right)$, but we have removed (2 red queen, 2 red jacks and 2 red kings) from a pack of cards .

Therefore total picture cards $\left( {12 - \left( {2 + 2 + 2} \right) = 12 - 6 = 6} \right)$

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{6}{{44}} = \dfrac{3}{{22}}$.

So, these are the required probabilities.

Note: Solving such probability questions simply requires understanding of what is being removed and how many cards are being left behind as the total sample cases is affected by this concept. Always remember there are only 3 face cards in each house and not 4 as ace is not a face card. We need to remember that there are in total 4 houses that are heart, diamond, and spade and club having13 cards in each of them.

Complete step-by-step answer:

In a deck of 52 cards we have 2 red queens, 2 red jacks, 2 red kings, 2 red aces one of hearts and one of diamond and we have removed them both so the total cards removed are

$\left( {2 + 2 + 2 + 2} \right) = 8$

So, the remaining cards are $\left( {52 - 8} \right) = 44$.

Therefore total outcomes $ = 44$

$\left( i \right)$ A black queen

In the remaining cards there are only 2 black queens, one of spade and one of club.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{2}{{44}} = \dfrac{1}{{22}}$.

$\left( {ii} \right)$ A red card

As we know that in a pack of 52 cards there are 4 houses having 13 cards each. (Spade, heart, diamond and club)

Therefore the total number of red cards is $\left( {13 + 13} \right) = 26$ (heart and diamond).

But we have removed 8 red cards, so the total number of red cards remaining are $\left( {26 - 8 = 18} \right)$.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{{18}}{{44}} = \dfrac{9}{{22}}$.

$\left( {iii} \right)$ A black jack

In the remaining cards there are only 2 black jacks, one of spade and one of club.

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{2}{{44}} = \dfrac{1}{{22}}$.

$\left( {iv} \right)$ A picture card

As we know in a pack of 52 cards (Jacks, kings and queens are picture cards), so, there are 4 jacks, 4 kings and 4 queens in a pack of 52 cards.

So the total picture card becomes $\left( {4 + 4 + 4 = 12} \right)$, but we have removed (2 red queen, 2 red jacks and 2 red kings) from a pack of cards .

Therefore total picture cards $\left( {12 - \left( {2 + 2 + 2} \right) = 12 - 6 = 6} \right)$

So, the probability becomes $ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{6}{{44}} = \dfrac{3}{{22}}$.

So, these are the required probabilities.

Note: Solving such probability questions simply requires understanding of what is being removed and how many cards are being left behind as the total sample cases is affected by this concept. Always remember there are only 3 face cards in each house and not 4 as ace is not a face card. We need to remember that there are in total 4 houses that are heart, diamond, and spade and club having13 cards in each of them.