Dave leaves his house and bikes directly east for 3 miles. He then turns and bikes directly
1 answer:
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Answer:
a) P(X∩Y) = 0.2
b) = 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is:
<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) =