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.
Y=2x+3
put x =0 onetime then y=3
now put y =0 2x+3
x= -3/2
Step-by-step explanation:
my answer is x-7
hmmm.. I hope its correct#
In this item, we are not given with the figure but knowing that these lines ought to be perpendicular then we will be able to derive the relationship between the slopes of the line.
The slopes of the perpendicular line are the negative reciprocals of one another. If we represent the slopes of the lines as m₁ and m₂, the relationship can be written in the form,
(m₁)(m₂) = -1
We are given with one of the slopes. To determine the value of the second slope then,
m₂ = -1/m₁
m₂ = -1/(2/5)
m₂ = -5/2
<em>ANSWER: m₂ = -5/2</em>
I'm pretty sure that this is a trick question, the answer is 61%.
