Could Someone give me a run down of how these problems are done? i used a calculator to get the answers but now i want to know h
ow to do these on my own
1 answer:
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Answer:
The probability is 0.02667
Step-by-step explanation:
Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.
So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:
P(D2/D1) = P(D2∩D1)/P(D1)
Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:
At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:
Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:
So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:
P(A)=1/3
P(B)=1/3
P(C)=1/3
Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:
P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)
P(D2/D1) = 0.02667
These array of numbers shown above are called matrices. These are rectangular arrays of number that are arranged in columns and rows. It is mostly useful in solving a system of linear equations. For example, you have these equations
x+3y=5
2x+y=1
x+y=10
In matrix form that would be
![\left[\begin{array}{ccc}1&3&5\\2&1&1\\1&1&10\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%265%5C%5C2%261%261%5C%5C1%261%2610%5Cend%7Barray%7D%5Cright%5D%20)
where the first column are the coefficients of x, the second column the coefficients of y and the third column is the constants, When you multiple matrices, just multiply the same number on the same column number and the same row number. For this problem, the solution is
![\left[\begin{array}{ccc}3*2&0*8\\2*0.6&-1*3\end{array}\right] = \left[\begin{array}{ccc}6&0\\1.2&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%2A2%260%2A8%5C%5C2%2A0.6%26-1%2A3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%5C%5C1.2%26-3%5Cend%7Barray%7D%5Cright%5D%20)
x+3y=5
2x+y=1
x+y=10
In matrix form that would be
where the first column are the coefficients of x, the second column the coefficients of y and the third column is the constants, When you multiple matrices, just multiply the same number on the same column number and the same row number. For this problem, the solution is