All categories

3 years ago

A card is chosen from a standard deck of cards. What is the probability that the card is a club, given that the card is black?

2 answers:

7 0

Answer : 13/52

There are 52 cards in a standard deck

13 of those cards are clubs, and they are all black.

13/52 will be a black club

leonid [27]3 years ago

5 0

From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26 A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace. P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13 WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces? P(AA) = (4/52)(3/51) = 1/221. 1 WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king? P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also 4/52 and the third is 4/52. We multiply these three individual probabilities together to get P(QQQ) = P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible. Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit) What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note: A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with 13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant, and you keep each card as it is dealt -- it's not returned to the deck.) The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for the other three. The probability of the royal flush is therefore the product of these numbers, or 5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154

You might be interested in

How do you solve this question and what’s the answer

shutvik [7]

Answer:

7.74 square cm

Step-by-step explanation:

To find the leftovers, calculate the area of the square and the area of the circle. Then subtract the two.

Square: A = s*s = 6*6 = 36

Circle: $A = \pi r^2\\A = \pi 3^2\\A = 9 \pi\\A= 28.26$

The area leftover is A = 36 - 28.26 = 7.74

4 0

3 years ago

If f(x)=12x+2(x-1), find f(6)

zavuch27 [327]

Answer:

12x6+2(6-1)

72+2x5

72+10

82 is the answer.

Step-by-step explanation:

Just replace the X's in the problem with 6.

5 0

3 years ago

determine the average rate of change of the function between the given variables h(x) = x; x=a, x=a+h

Alecsey [184]

Answer:

the average rate of change is 4.

Step-by-step explanation:

Find the average rate of change of f(x)=x^2 on the interval [1,3].

The average rate of change of f(x) on the interval [a,b] is f(b)−f(a)/b−a.

We have that a=1, b=3, f(x)=x^2.

Thus, f(b)−f(a)/b−a=((3))^2−(((1))^2)/3−(1) = 4.

6 0

3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago

Write the equation of a line, in SLOPE INTERCEPT FORM, that is perpendicular to the given equation and passes through the given

adoni [48]

Step-by-step explanation:

slope interception formula is

y-y1=m(x-x1)

where m is m=y-y1/x-x1 in this case m=-2 because the line we are trying to find is parallel to the given one y=-2x-6 where slope k=-2

so the final equation would be

y-1=-2(x-(-4))

y-1=-2x-2*4

y=-2x-8+1=-2x-7

4 0

3 years ago