An urn contains two blue balls (denoted B1 and B2) and three white balls (denoted W1, W2, and W3). One ball is drawn, its color
1 answer:
Answer:
(a) As shown below.
(b) The likelihood that the first ball drawn is blue is 0.40.
(c) The chance that only white balls are drawn is 0.36.
Step-by-step explanation:
The balls present in the urn include:
Blue balls: B₁ and B₂
White balls: W₁, W₂ and W₃
It is stated that two balls will be drawn with replacement, and their colors noted.
(a)
The potential outcomes for selecting two balls are listed as follows:
B₁B₁ B₂B₁ W₁B₁ W₂B₁ W₃B₁
B₁B₂ B₂B₂ W₁B₂ W₂B₂ W₃B₂
B₁W₁ B₂W₁ W₁W₁ W₂W₁ W₃W₁
B₁W₂ B₂W₂ W₁W₂ W₂W₂ W₃W₂
B₁W₃ B₂W₃ W₁W₃ W₂W₃ W₃W₃
There are a total of N = 25 possible outcomes.
(b)
The sample space for drawing a blue ball first is:
S = {B₁B₁, B₁B₂, B₁W₁, B₁W₂, B₁W₃, B₂B₁, B₂B₂, B₂W₁, B₂W₂, B₂W₃}
n(S) = 10
Hence, the probability of the first ball being blue is calculated as follows:
Thus, the probability of the first ball drawn being blue is 0.40.
(c)
The sample space for drawing only white balls is:
X = {W₁W₁, W₂W₁, W₃W₁, W₁W₂, W₂W₂, W₃W₂, W₁W₃, W₂W₃, W₃W₃}
n(X) = 9
The probability that only white balls are drawn can be computed as such:
Thus, the probability that only white balls are drawn is 0.36.
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Step-by-step breakdown:
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To achieve a positive profit p(x)>0, we start with:
Next, we solve for x:
The parameters are:
a = -2
b = 7
c = -3
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b = 7
c = -3
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This yields two results—one a fraction, the other a whole number. We focus on the whole number since the muffins sold must be a complete unit. Thus we conclude:
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