Question

1 point
There are n bins of which the kth contains k − 1 blue balls and n − k red balls. You pick a bin at random and remove two balls at random without replacement. Find the probability that:

• the second ball is red;
• the second ball is red, given that the first is red.
a) 1/3, 2/3
b) 1/2, 1/3
c) 1/2, 2/3
D) 1/3, 1/3​

Answer 1

Answer:

Sol (c) 1/2, 1/3

Step-by-step explanation:

Let Ci be the colour of the ith ball. In each bin, there are a total of (k −1) + (n−k) = (n−1)

balls. Of these half are blue and the other half are red (verify Pn

k=1 k − 1 = Pn

k=1 n − k).

The probability of the second ball being red is equal to the probability of the second ball being

red given that the first ball was either red or blue.

For a particular bin we have,

P(C2 = red) = (n − k)

(n − 1)

(n − k − 1)

(n − 2) +

(k − 1)

(n − 1)

(n − k)

(n − 2) =

n − k

n − 1

Considering all bins, we have

P(C2 = red) = Xn

k=1

n − k

n(n − 1) =

1

2

The probability of the second ball being red given that the first ball was red,

P(C2 = red|C1 = red) = P(C2 = red, C1 = red)

P(C1 = red)

Now, P(C1 = red) = 1/2

For a particular bin,

P(C2 = red, C1 = red) = (n − k)

(n − 1)

(n − k − 1)

(n − 2)

Considering all bins, we have

P(C2 = red|C1 = red) = Xn

k=1

(n−k)(n−k−1)

n(n−1)(n−2)

1

2

Simplifying, we have

P(C2 = red|C1 = red) = 2/3

Answer 2

Given:

• There are n bins, where the kth bin contains k-1 blue balls and n-k red balls.
• Two balls are picked at random from a randomly chosen bin, without replacement.

To Find:

• The probability that the second ball is red.
• The probability that the second ball is red, given that the first ball is red.

Solution:

Probability that the second ball is red:

• Let R1 be the event that the first ball is red, and R2 be the event that the second ball is red.
• Then, using the law of total probability, we have P(R2) = Σk P(R2|k)P(k), where P(k) is the probability of choosing the kth bin and P(R2|k) is the probability of the second ball being red given that the bin is k.
• P(k) = 1/n for each k, since the bin is chosen at random.
• P(R2|k) = (n-k)/(2n-k-1), since there are n-k red balls and 2n-k-1 total balls in the kth bin.
• Therefore, P(R2) = Σk (n-k)/(2n-k-1) * (1/n).
• Simplifying this expression gives P(R2) = 1/2, regardless of the value of n.

Probability that the second ball is red, given that the first ball is red:

• Let R1 be the event that the first ball is red, and R2 be the event that the second ball is red.
• Then, using Bayes' theorem, we have P(R2|R1) = P(R1 ∩ R2) / P(R1), where P(R1 ∩ R2) is the probability that both balls are red and P(R1) is the probability that the first ball is red.
• P(R1) = Σk (n-k)/(2n-k-1) * ((k-1)/n), since the first ball must be red and the bin must be k.
• P(R1 ∩ R2) = Σk [(n-k)/(2n-k-1)] * [(k-1)/(n-1)], since the first ball must be red and the second ball must be red and in the same bin as the first.
• Simplifying these expressions gives P(R1) = (n-1)/(2n-1) and P(R1 ∩ R2) = (n-2)/(4n-6), regardless of the value of n.
• Therefore, P(R2|R1) = P(R1 ∩ R2) / P(R1) = (n-2)/(4n-6) * (2n-1)/(n-1) = 2/3, regardless of the value of n.

Therefore, the correct option is (c) 1/2, 2/3.

To learn more about probability from the given link.

https://brainly.in/question/996463

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