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# from a pack of 52 cards, one card is drawn at random. evaluate probability that the card drawn is a nine or a heart?

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## A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen

Click here👆to get an answer to your question ✍️ A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen

A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is

Question

(i) a black king   (ii) either a black card or a king  (iii) a jack, queen or a king   (iv) neither an ace nor a king   (v) spade or an ace   (vi) neither a red card nor a queen   (vii) other than an ace   (viii) a ten   (ix) a spade   (x) a black card   (xi) the seven of clubs   (xii) jack   (xiii) the ace of spades   (xiv) a queen    (xv) a heart     (xvi) a red card  (xvii) neither a king nor a queen

No. of cards in a pack =52

Medium Open in App Solution Verified by Toppr

Solution(i):

No. of black kings =2

Therefore, 2 C 1 ​

( Selecting 1 out of 2 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) a black king is picked.

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

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 2 C 1 ​ ​ = 52 2 ​ = 26 1 ​

Solution(ii):

No. of black cards or kings =28..... (26Black(including 2 Black kings) + 2 Red Kings)

Therefore, 28 C 1 ​

( Selecting 1 out of 28 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) a either a black card or a king is picked.

Let E be the event of getting either a black card or a king from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 28 C 1 ​ ​ = 52 28 ​ = 13 7 ​

Solution(iii):

No. of jack, queen or king =12 (4- Jack, 4- Queen, 4-King)

Therefore, 12 C 1 ​

( Selecting 1 out of 12 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) a jack, queen or a king is picked.

Let E be the event of getting a jack, queen or a king from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 12 C 1 ​ ​ = 52 12 ​ = 13 3 ​

Solution(iv):

Therefore, 16 C 1 ​

( Selecting 1 out of 16 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) a spade or an ace is picked.

Let E be the event of getting a spade or an ace from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 16 C 1 ​ ​ = 52 16 ​ = 13 4 ​

Solution(v):

No. of neither ace nor king =52−8=44 ...... (As there are 4-Kings and 4-Aces)

Therefore, 44 C 1 ​

( Selecting 1 out of 44 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) a neither an ace nor a king is picked.

Let E be the event of getting neither an ace nor a king from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 44 C 1 ​ ​ = 52 44 ​ = 13 11 ​

Solution(vi):

No. of neither red nor queen \$\$=52-28=24\$\$  ...(26-red+ 2-Black Queen)

Therefore, 24 C 1 ​

( Selecting 1 out of 24 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) neither red nor a queen is picked.

Let E be the event of getting neither red nor queen from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 24 C 1 ​ ​ = 52 24 ​ = 13 6 ​

Solution(vii):

No. of non-ace cards =52−4= 48 ......(4 Aces)

Therefore, 48 C 1 ​

( Selecting 1 out of 48 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) non-ace card is picked.

Let E be the event of getting a non-ace card from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 48 C 1 ​ ​ = 52 48 ​ = 13 12 ​

Solution(viii):

No. of cards with no. 10 =4

Therefore, 4 C 1 ​

( Selecting 1 out of 4 items) times out of

52 C 1 ​

( Selecting 1 out of 52 items) card with no. 10 is picked.

Let E be the event of getting a card with no. 10 from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

​ = 52 C 1 ​ 4

स्रोत : www.toppr.com

## From a pack of 52 cards, one card is drawn in random. What is the probability of the card drawn is a 10 or a spade?

Answer (1 of 6): There are 4 Number 10 cards in a deck of 52 cards. There are 13 cards each of 4 suits (Spades♠, Hearts♥, Clubs♣ & Diamonds♦) So, number of Spade♠ cards = 13 But, out of the 13 Spade cards, 1 is a number 10 card. We cannot count a card two times. Hence, the number of desired ...

From a pack of 52 cards, one card is drawn in random. What is the probability of the card drawn is a 10 or a spade?

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When I see this sort of auestion, I always wonder whether it is badly-worded on purpose or whether it wasn't peer-reviewed properly!

From a pack of 52 cards one card is drawn in random. What is the Probability of the card drawn is a 10 or a spade?

Question: Hopefully the writer meant AT random not IN random.

Question: Does the writer mean What is the Probability that the card drawn is EITHER a 10 or

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Let S S

denote the event that it is a spade and let

E E

denote the event that it is a 10.

Then

P(S∪E)=P(S)+P(E)−P(S∩E)=

1 4 + 1 13 − 1 52 = 16 52

P(S∪E)=P(S)+P(E)−P(S∩E)=14+113−152=1652

Kurusar Iyer

Online gaming & card games enthusiast, specially affiliated to RummyPassion.comAuthor has 188 answers and 1.4M answer views3y

There are 4 Number 10 cards in a deck of 52 cards.

There are 13 cards each of 4 suits (Spades♠, Hearts♥, Clubs♣ & Diamonds♦)

So, number of Spade♠ cards = 13

But, out of the 13 Spade cards, 1 is a number 10 card.

We cannot count a card two times.

Hence, the number of desired cards= 13 + 4 - 1 = 16

Probability of drawing a 10 or a Spade = 16/52 = 4/13

For more clarity, have a look at a standard deck of cards:

Count and see for yourself.

It will appear simple!

Edward Sherry

Worked at Expert Research AssociatesAuthor has 4.8K answers and 3M answer views4y

There are four tens and 13 spades, but one of the spades is the ten of spades, and we don’t want to double count. So 16 cards qualify as either a 10 or a spade (or both). The probability of drawing a 10 or a spade is 16/52 = 4/13.

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Th possibility the drawn card is a 10 is 4/52. or 1/13. The possibility it is a spade is 1/4.

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There are 13 spades in a standard deck of cards; and 4 10’s, which gives us a total of 17 cards. However, one of those cards is the 10 of spades, so we subtract that one card, which keeps us from counting it twice.

Thus, we have 16 out of 52, or 16/52. And 16/52=4/13 as the probability that one card, chosen at random, will be either a spade or a 10, or the 10 of spades.

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स्रोत : www.quora.com

## A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is,(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen.

A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is,(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red ...

A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is,

(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen.

Answer Verified 169k+ views 2 likes

Hint: There are 52 cards in a deck. The probability of finding a card can be calculated by dividing the number of cards of the given type by the total number of cards.

Total number of cards (T) = 52

(i) a black king:

In a deck of cards, there are two black kings, one of spade and one of clubs.

∴ ∴

Probability of finding a black king

= 2 25 = 1 13 =225=113

(ii) either a black card or a king:

Total black cards = 13 x 2 = 26

Total kings other than black cards = 2

∴ ∴ Probability = 2×26 52 = 28 52 = 7 13 =2×2652=2852=713

(iii) a jack, a queen or a king:

Total jacks = 4 Total queens = 4 Total kings = 4 ∴ ∴ Probability = 4+4+4 52 = 12 52 = 3 13 =4+4+452=1252=313

(iv) neither an ace nor a king:

The required cards are all cards other than kings and aces.

⇒(T)−(number of kings+number of aces)

⇒(T)−(number of kings+number of aces)

Number of kings = 4 Number of aces = 4 ∴ ∴ Probability = (T)−(4+4) 52 = 52−8 52 = 44 52 = 11 13

=(T)−(4+4)52=52−852=4452=1113

(v) a spade or an ace

Number of cards of spades = 13

Aces other than spades = 4 – 1 = 3

∴ ∴ Probability = 3 13 =313

(vi) neither a red card nor a queen

Required cards are all cards other than red cards and queens.

Red cards = 13 x 2 = 26

Queens other than those included in red cards = 4 – 2 = 2

Therefore, probability

= T−(26+2) 52 =T−(26+2)52 = 52−28 52 = 24 52 = 6 13 =52−2852=2452=613

(vii) other than an ace

Required cards are all cards other than ace.

⇒T−(number of aces) ⇒T−(number of aces) Number of aces = 4 ∴ ∴ Probability = T−4 52 = 52−4 52 = 48 52 = 12 13

=T−452=52−452=4852=1213

(viii) a ten Number of tens = 4 ∴ ∴ Probability = 4 52 = 1 13 =452=113 (ix) a spade

∴ ∴ Probability = 13 52 = 1 4 =1352=14 (x) a black card

Number of black cards = number of spades + number of clubs

= 13 + 13 = 26 ∴ ∴ Probability = 26 52 = 1 2 =2652=12

(xi) the seven of clubs

There are only one seven clubs in one deck of cards.

∴ ∴ Probability = 1 52 =152 (xii) a jack Number of jacks = 4 ∴ ∴ Probability = 4 52 = 1 13 =452=113

There is only one ace of spades in a deck of cards.

∴ ∴ Probability = 1 52 =152 (xiv) a queen

Number of queens = 4

∴ ∴ Probability = 4 52 = 1 13 =452=113 (xv) a heart

Number of hearts = 13

∴ ∴ Probability = 13 52 = 1 4 =1352=14 (xvi) a red card

Number of red cards = number of hearts + number of diamonds

= 13 + 13 =26 ∴ ∴ Probability = 26 52 =2652

(xvii) neither a king nor a queen

The required cards are all cards except kings and queens.

⇒T−(number of kings + number of queens)

⇒T−(4+4)

⇒T−(number of kings + number of queens)⇒T−(4+4)

∴ ∴ Probability = 52−8 52 = 44 52 = 11 13 =52−852=4452=1113

Note: (1) Make sure to not count a card twice like in part (ii) or (v).

(2) Ace is not a face card. Many students make that mistake.

(3) This is the distribution of a deck of playing cards:

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each; i.e. spades, hearts, diamonds and clubs. Cards of spades and clubs are black cards. Cards of hearts and diamonds are red cards. The cards in each suit are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.

स्रोत : www.vedantu.com

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Mohammed 8 day ago

Guys, does anyone know the answer?