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.
Answer:
Answer:
d
Step-by-step explanation:
The number of years must be non-negative.
This eliminates all of the options except for d.
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I believe the 11/25 one would be 44% and 4/50 would be 8%, 4/5 would be 80%. Hope it helps!
Complete question :
Match the numbers in scientific notation with the corresponding numbers in standard notation.
232,000,000
0.00000431
2,320,000,000
0.000000000000431
2.32E9
2.32E8
4.31E-13
4.31E-6
Answer:
2.32E8 = 232,000,000
2.32E9 = 2,320,000,000
Step-by-step explanation:
2.32E8 - - > 8 digits after the decimal point ; 232,000,000
2.32E9 - - > 9 digits after the decimal point ;
2,320,000,000
4.31E-13 - - > 13 preceeding 0's before digit '4' ; 0.000000000000431
4.31E-6 - - > 6 preceeding 0's before digit '4' ; 0.000000000000431
