Answer:
A darts player practices throwing a dart at the bull’s eye on a dart board. Her probability of hitting the bull’s eye for each throw is 0.2.
(a) Find the probability that she is successful for the first time on the third throw:
The number F of unsuccessful throws till the first bull’s eye follows a geometric
distribution with probability of success q = 0.2 and probability of failure p = 0.8.
If the first bull’s eye is on the third throw, there must be two failures:
P(F = 2) = p
2
q = (0.8)2
(0.2) = 0.128.
(b) Find the probability that she will have at least three failures before her first
success.
We want the probability of F ≥ 3. This can be found in two ways:
P(F ≥ 3) = P(F = 3) + P(F = 4) + P(F = 5) + P(F = 6) + . . .
= p
3
q + p
4
q + p
5
q + p
6
q + . . . (geometric series with ratio p)
=
p
3
q
1 − p
=
(0.8)3
(0.2)
1 − 0.8
= (0.8)3 = 0.512.
Alternatively,
P(F ≥ 3) = 1 − (P(F = 0) + P(F = 1) + P(F = 2))
= 1 − (q + pq + p
2
q)
= 1 − (0.2)(1 + 0.8 + (0.8)2
)
= 1 − 0.488 = 0.512.
(c) How many throws on average will fail before she hits bull’s eye?
Since p = 0.8 and q = 0.2, the expected number of failures before the first success
is
E[F] = p
q
=
0.8
0.2
= 4.
The domain is the x-value of the line
Therefore since the point to the left is not filled in and the point to the right is filled in then your answer would be -11
9514 1404 393
Answer:
Step-by-step explanation:
The left section is defined at x = -5, but not at x = 0, so the interval for that definition is -5 ≤ x < 0. The function value on that interval is constant: 3.
The right section is defined a x = 5, but not at x = 0, so the interval for that definition is 0 < x ≤ 5. That section is a straight line with a slope of -1 and a y-intercept of 1. The function value on that interval can be written -x+1.
Then the piecewise definition is ...
f(x) = {3, for -5 ≤ x < 0; -x+1, for 0 < x ≤ 5}
250.35 that the anwers hope this help a lot
