An urn contains 20 red balls and 40 blue balls. two are chosen at random, one after the other, without replacement. (round your answers to one decimal place.) (a) use a tree diagram to calculate the following probabilities: the probability (as a %) that both balls are red 33 x % the probability (as a %) that the first ball is red and the second is not 33 x % the probability (as a %) that the first ball is not red and the second is red the probability (as a %) that neither ball is red (b) what is the probability (as a %) that the second ball is red? % (c) what is the probability (as a %) that at least one of the balls is red?

a) Both balls red: P = 10.73%

Red then blue:  P = 22.6%.

Blue then red: P = 22.6%

Both blue: P = 44%

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b) P = 33.3%

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c) P = 56%

Step-by-step explanation:

a)The probability that both balls are red.

Initially, there are 60 total balls, 20 of which are red.

So, P1, which is the probability that the first ball is red is

[tex]P1 = \frac{number of red balls}{number of total balls} = \frac{20}{60} = 1/3 = 0.333[\tex]

Considering there are no replacement, there are now 59 balls, 19 of which are red. The probability of the second ball being red is

[tex]P2 = \frac{19}{59} = 0.322[\tex]

The probability of both balls being red is P = P1*P2 = 0.333*0.322 = 0.1073 = 10.73%.

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The probability of the first ball being red has already been calculated, it is P1 = 0.333. For the probability of the second ball being blue, there are 59 balls, 40 of which are blue. So, the probability of the second ball being blue is

[tex]P2 = \frac{40}{59} = 0.68[\tex]

The probability of the first ball being red and the second blue is P = P1*P2 =  0.333*0.68 = 0.226 = 22.6%.

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The probability that the first ball being blue is

[tex]P1 = \frac{number of blue balls}{total number of balls} = \frac{40}{60} = 2/3 = 0.6666[\tex]

There are now 59 balls, 20 of which are red. So the probability P2 of the second ball being red is

[tex]P2 = \frac{20}{59} = 0.34[\tex]

So, the probability of choosing a blue ball then a red ball is P = P1*P2 = 0.666*0.34 = 0.226 = 22.6%

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In this case, the desired outcome is both balls being blue.

The probability P1 of the first ball being blue is 2/3 = 0.6666.

There are now 59 balls, 39 of which are blue.

The probability P2 of the second ball being blue is 39/59 = 0.66

So, the probability of both balls being blue is P = 0.6666*0.666 = 0.44 = 44%

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b) There are two cases in which the second ball is red. The first case is when the first ball is red, and the second case is when the first ball is blue.

The probability of the second ball being red is P = P1+P2, where P1 is the probability of the sequence being red-red, and P2 is the probability of the sequence being blue-red. From a), we have P1 = 0.1073 and P2 = 0.226. So P = 0.1073 + 0.226 = 33,3%.

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c) The sum of total probabilities are 1. So the probability P of at least one ball being red can be formulated as P = 1-Pbb, where Pbb is the probability of both balls being blue. From a), Pbb = 44%. So P = 1-0.44 = 0.56 = 56%.

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

i believe the answer is c.