Respuesta :
Answer:
a) Figure attached
b) [tex] P(DD) =0.1*0.1= 0.01[/tex]
Step-by-step explanation:
For this case we can find first the probability of being dfective on the shipment like this:
[tex]P(D) = \frac{1000}{10000}=0.1[/tex]
Since he select 10000 and 1000 of theme were defective.
Now we know that the inspector select 2 switches at random so then we can create the possible outcomes.
Part a
The tree diagram is on the figure attached, the sample space on this case is:
[tex] S=[DN , DD, NN, ND][/tex]
D = represent a defective switch
N= represent a non defective swicth
Part b
For this case we want the probability that both switches would be defective and on this case we can do this:
[tex] P(DD) =0.1*0.1= 0.01[/tex]
Assuming independent between the events.
The probability that both switches are defective is 1%.
Given that a shipment contains 10,000 switches, and of these, 1000 are bad, and an inspector draws 2 switches at random, one after the other, to find the probability that both switches are defective the following calculation must be performed:
- 1000/10000 = 1/10
- 1/10 x 1/10 = X
- 0.1 x 0.1 = X
- 0.01 = X
- 0.01 x 100 = 1
Therefore, the probability that both switches are defective is 1%.
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