Astandard deck of playing cards is shuffled and three people each choose a card. find the probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement, and if the cards are chosen without replacement. (a) the cards are chosen with replacement. (b) the cards are chosen without replacement.

Part (a) The cards are chosen with replacement is 0.03125 or 1/32

Part (b) The cards are chosen without replacement is 0.0306 or 13/425.

Step-by-step explanation:

Consider the provided information.

A standard deck of playing cards is shuffled 3 cards are chosen.

In a deck of card 13 are spade out of 52 cards. 26 cards are red out of 52 cards.

We need to find the probability that the first two cards chosen are spades and the third card is red.

Part (a) The cards are chosen with replacement.

Suppose the first card is spade then the probability is P(A) = 13/52

As spade are 13 out of 52.

Now, the card is replaced back in the deck and we again draw a card. The probability of getting spade in second time is P(B) = 13/52.

The card is replaced back in the deck and we again draw a card. The probability of getting a red card in third time is P(C) = 26/52.

Now Multiply 3 fraction together as shown.

Hence, the probability of getting the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 or 1/32.

Part (b) The cards are chosen without replacement.

Suppose the first card is spade then the probability is P(A) = 13/52

Now, the card is not replaced back in the deck and we again draw a card. The probability of getting spade in second time is P(B) = 12/51.

The probability of getting a red card in third time is P(C) = 26/50.

Now Multiply 3 fraction together as shown.

Hence, the probability of getting the first tow cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.0306 or 13/425.

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step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).