There could be 25 bags with 2 blueberry scones in each and 3 cranberry scones in each. This is because 25 is a factor of both amounts so you would divide each by 25. That leaves it with 2 blueberry scones and 3 cranberry scones.
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Answer:</h2>
The probability that a defective component came from shipment II is:
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Step-by-step explanation:</h2>
Let A denote the event that the defective component was from shipment I
Also, P(A)=2%=0.02
and B denote the event that the defective component was from shipment II.
i.e. P(B)=5%=0.05
Also, P(shipment I is chosen)=1/2=0.5
and P(shipment II is chosen)=1/2=0.5
The probability that a defective component came from shipment II is calculated by Baye's rule as follows:
Hence, the answer is:
Answer:
dy/dx = -1/√(1 - x²)
For 0 < y < π
Step-by-step explanation:
Given the function cos y = x
-siny dy = dx
-siny dy/dx = 1
dy/dx = -1/siny (equation 1)
But cos²y + sin²y = 1
=> sin²y = 1 - cos²y
=> siny = √(1 - cos²y) (equation 2)
Again, we know that
cosy = x
=> cos²y = x² (equation 3)
Using (equation 3) in (equation 2), we have
siny = √(1 - x²) (equation 4)
Finally, using (equation 4) in (equation 1), we have
dy/dx = -1/√(1 - x²)
The largest interval is when
√(1 - x²) = 0
=> 1 - x² = 0
=> x² = 1
=> x = ±1
So, the interval is
-1 < x < 1
arccos(1) < y < arxcos(-1)
= 0 < y < π
Answer:
<em>Find the probability of success in a single trial and then think about the nature of the problem (when do we stop). </em>
Step-by-step explanation:
Observe that in the single trial, we have (8 4) possibilities of choosing our set of balls. If we have chosen two white balls and two black balls, the probability of doing that is simply
p=(4 2)*(4 2)/(8 4)
This is well know Hyper geometric distribution. Now, define random variable X that marks the number of trials that have been needed to obtain the right combination (two white and two black balls). From the nature of the problem, observe that X has Geometric distribution with parameter p that has been calculated above. Hence
P(X = n) = (1— p)^n-1 *( p )
<em>Find the probability of success in a single trial and then think about the nature of the problem (when do we stop). </em>
Answer:
The probability of you winning is 1/3
No, it is not a fair game
Step-by-step explanation:
The first thing we need to have here is the sample space. This refers to the set of all possible results that can occur from the rolling.
Please check attachment for this
Kindly note that the total number of possible outcomes is 36.
And in the attachment, sums which are divisible by 3 are circled.
The number of circles we can count is 12
Thus, the probability of you winning would be number of circles/total number of outcomes = 12/36 = 1/3
Is it a fair game?
No, it is not
It can only be a fair game if the probability of winning equals probability of losing ( which is 18/36 = 1/2 or 0.5)
