Question

Four cards are drawn out from a packet of cards casually. What is the probability to find 2 cards of them to be spade? @probability

Answer 1

`17160/6497400`

Explanation:

There are 52 cards altogether, and 13 of them are spades.

Probability of drawing the first spade is:

`13/52`

Probability of drawing a second spade is:

`12/51`

This is because, when we have picked out the spade, there are only 12 spades left and consequently only 51 cards altogether.

probability of drawing a third spade:

`11/50`

probability of drawing a fourth spade:

`10/49`

We need to multiply all these together, to get the probability of drawing a spade one after another:

`13/52*12/51*11/50*10/49=17160/6497400`

So the probability of drawing four spades concurrently without replacement is:

`17160/6497400`

Answer 2

`(57,798)/(270,725)~~21.35%`

Explanation:

Let's first see the number of ways we can pick 4 cards from a pack of 52:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

`C_(52,4)=(52!)/((4!)(48!))=(52xx52xx50xx49)/24=270,725`

How many ways can we draw 4 cards and have exactly 2 of them be spades? We can find that by choosing 2 from the population of 13 spades, then choosing 2 cards from the remaining 39 cards:

`C_(13,2)xxC_(39,2)=(13!)/((2!)(11!))xx(39!)/((2!)(37!))=(13xx12)/2xx(39xx38)/2=57,798`

This means the probability of drawing exactly 2 spades on a 4 card draw from a standard deck is:

`(57,798)/(270,725)~~21.35%`

Answer 3

`0.21349 = 21.349 %`

Explanation:

`C_2^4 (13/52)(12/51)(39/50)(38/49)`
`= ((4!)/(2!2!)) (1/4)(17784/124950)`
`= (6/4)(17784/124950)`
`= 4446/20825`
`= 0.21349`
`= 21.349 %`

`"Explanation : "`
`"We express that the first and second card has to be a spade."`
`"Then the third and fourth card cannot be a spade. Of course"`
`"the spades could be in another place, like 2nd and 4th and so"`
`"on so therefore we multiply by "C_2^4"."`
`"First draw : there are 13 spades cards on 52 "=> 13/52`
`"2nd draw : there are 12 spades cards left on 51 cards "=> 12/51`
`"3th draw : 39 non-spades cards left on 50 cards "=> 39/50`
`"4th draw : 38 non-spades cards left on 49 cards "=> 38/49`

Answer 4

The probability is approximately `21.35%`.

Explanation:

Visualize the deck in two parts: the spades, and everything else.

The probability we seek is the number of hands with two cards from the spades and two cards from everything else, divided by the number of hands with any 4-cards.

Number of hands with 2 spades and 2 non-spades: From the 13 spades, we will choose 2; from the other 39 cards, we will choose the remaining 2. The number of hands is `""_13C_2 xx ""_39C_2.`

Number of hands with any 4 cards: From all 52 cards, we will choose 4. The number of hands is `""_52C_4.`

`"P"("2 spades out of 4") = [((13),(2))((39),(2))]/[((52),(4))] = (""_13C_2 xx ""_39C_2)/(""_52C_4)`

Notice that the 13 and 39 in the top row add to the 52 in the bottom row; same with 2 and 2 adding to 4.

`"P"("2 spades out of 4") = [" "(13xx12)/(2xx1)xx(39xx38)/(2xx1)" "]/[(52xx51xx50xx49)/(4xx3xx2xx1)]`

`color(white)("P"("2 spades out of 4")) = [(13xx6)xx(39xx19)]/(13xx17xx25xx49)`

`color(white)("P"("2 spades out of 4")) = [6xx39xx19]/(17xx25xx49)`

`color(white)("P"("2 spades out of 4")) = "4,446"/"20,825" " "~~21.35%`

In general, any probability question that divides a "population" (like a deck of cards) into a few distinct "sub-populations" (like spades vs. other suits) can be answered in this way.