The answer is no solution
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:

<span>100 is not divisible by 3 when u divide 100 by 3 u get 33.333333...... not 3.33333.....
the good explanation of your answer is that 0.999999....is equal to 1
we can prove it by the following way:
let x=0.9999....
then 10x = 9.999999.....
subtract we will get 10x - x = 9.9999....- 0.99999......
which gives us 9x = 9 and therefore x =1
i hope this will help u with your confusion....... :)</span><span>
</span>
Answer:
9,020,406.27
Step-by-step explanation: