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<u>Answer:</u>
The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744
<u>Solution:</u>
Total number of coils = number of good coils + defective coils = 88 + 12 = 100
p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) p(getting 2 good coils for second selection)
p(first selection) = p(second selection) =
Hence, p(getting 2 good coil for two selection) =
Answer:
(1) The possible outcomes are: X = {0, 1, 2, 3}.
(2) The number of times should Hartley spin a difference of 1 is 36.
(3) The number of times should Hartley spin a difference of 0 is 24.
Step-by-step explanation:
The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.
The total number of spins for each of the spinner is, <em>n</em> = 96.
(1)
The sample space of spinning both the spinners together are:
S = {(1, 1), (1, 2), (1, 3), (1, 4)
(2, 1), (2, 2), (2, 3), (2, 4)
(3, 1), (3, 2), (3, 3), (3, 4)
(4, 1), (4, 2), (4, 3), (4, 4)}
Total = 16.
The possible outcomes are:
X = {0, 1, 2, 3}.
(2)
The sample space with the difference 1 are:
S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}
n (S₁) = 6
The probability of the difference 1 is:
The spinners were spinner 96 times.
The expected number of times would Hartley spin a difference of 1 is:
Thus, the number of times should Hartley spin a difference of 1 is 36.
(3)
The sample space with the difference 0 are:
S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}
n (S₂) = 4
The probability of the difference 0 is:
The spinners were spinner 96 times.
The expected number of times would Hartley spin a difference of 0 is:
Thus, the number of times should Hartley spin a difference of 0 is 24.