Transcript Probability – tree diagrams

A bag contains 5 black and 3 yellow beads. A bead is selected at
random from the bag and retained. A second bead is then selected
from the remaining beads.
Find the probability that both beads were
(i)
black
(ii)
the same colour
Multiply ALONG
Branches
4/
B
BB
3/
Y
BY If you use more
7
B
5/
8
7
3/
8
5/
7
B
than 1 of the
YB outcomes ADD
2/
7
Y
YY
Y
A bag contains 5 black and 3 yellow beads. A bead is selected at
random from the bag and retained. A second bead is then selected
from the remaining beads.
Find the probability that both beads were
(i)
black
(ii)
the same colour
4/
B
BB
3/
Y
BY
5/
7
B
YB
2/
7
Y
YY
7
B
5/
8
7
3/
8
Y
𝟓
𝟖
×
𝟒
𝟕
=
𝟐𝟎
𝟓𝟔
A bag contains 5 black and 3 yellow beads. A bead is selected at
random from the bag and retained. A second bead is then selected
from the remaining beads.
Find the probability that both beads were
(i)
black
(ii)
the same colour
4/
B
BB
3/
Y
BY
5/
7
B
YB
7
B
5/
8
7
3/
8
=
𝟐𝟎
𝟓𝟔
+
=
𝟐𝟎
𝟓𝟔
𝟔
𝟓𝟔
Y
=
𝟐𝟔
𝟓𝟔
2/
7
Y
YY
=
𝟔
𝟓𝟔
A teacher walks, cycles or drives to school with probabilities of 0.1, 0.3 and 0.6 respectively.
If she walks to school, she has a probability of 0.35 of being late. The corresponding
probabilities of being late if she cycles or drives to school are 0.1 and 0.55 respectively.
(a)
Find the probability that she is late on any particular day.
(b)
Given that she is late one day, find the probability that she walked.
(c)
Given that she is not late one day, find the probability that she walked.
Give your answers to 3 decimal places.
0.1
0.3
0.6
WL
0.35
L
0.65
0.1
L’ WL’
L CL
0.9
0.55
L’
L
CL’
DL
0.45
L’
DL’
W
C
D
A teacher walks, cycles or drives to school with probabilities of 0.1, 0.3 and 0.6 respectively.
If she walks to school, she has a probability of 0.35 of being late. The corresponding
probabilities of being late if she cycles or drives to school are 0.1 and 0.55 respectively.
(a)
Find the probability that she is late on any particular day.
Give your answers to 3 decimal places.
P(L) = 0.035 + 0.03 + 0.33 = 0.395
0.1
0.3
0.6
WL
0.35
L
0.65
0.1
L’ WL’
L CL
0.9
0.55
L’
L
CL’
DL
0.45
L’
DL’
W
C
D
0.035
0.03
0.33
A teacher walks, cycles or drives to school with probabilities of 0.1, 0.3 and 0.6 respectively.
If she walks to school, she has a probability of 0.35 of being late. The corresponding
probabilities of being late if she cycles or drives to school are 0.1 and 0.55 respectively.
(b)
Given that she is late one day, find the probability that she walked.
Give your answers to 3 decimal places.
0.035
0.395
= 0.089 (3dp)
0.1
0.3
0.6
WL
0.35
L
0.65
0.1
L’ WL’
L CL
0.9
0.55
L’
L
CL’
DL
0.45
L’
DL’
W
C
D
0.035
0.03
0.33
A teacher walks, cycles or drives to school with probabilities of 0.1, 0.3 and 0.6 respectively.
If she walks to school, she has a probability of 0.35 of being late. The corresponding
probabilities of being late if she cycles or drives to school are 0.1 and 0.55 respectively.
(c)
Given that she is not late one day, find the probability that she walked.
Give your answers to 3 decimal places.
0.065
= 0.107 (3dp)
1 – 0.395
0.1
0.3
0.6
WL
0.35
L
0.65
0.1
L’ WL’
L CL
0.9
0.55
L’
L
CL’
DL
0.45
L’
DL’
W
C
D
A certain medical disease occurs in 1% of the population. A simple screening procedure is
available and in 8 out of 10 cases where the patient has the disease, it produces a positive
result. If the patient does not have the disease there is still a 0.05 chance that the test will
give a positive result.
Find the probability that a randomly selected individual:
(a)
does not have the disease but gives a positive result in the screening test
(b)
gives a positive result in the test.
Ann has taken the test and her result is positive.
(c)
Find the probability that she has the disease.
0.01
0.99
0.8
P
DP
0.2
P’
DP’
0.05
P
D’P
0.95
P’
D’P’
D
D’
A certain medical disease occurs in 1% of the population. A simple screening procedure is
available and in 8 out of 10 cases where the patient has the disease, it produces a positive
result. If the patient does not have the disease there is still a 0.05 chance that the test will
give a positive result.
Find the probability that a randomly selected individual:
(a)
does not have the disease but gives a positive result in the screening test
= 0.99 x 0.05 = 0.0495
0.01
0.99
0.8
P
DP
0.2
P’
DP’
0.05
P
D’P
0.95
P’
D’P’
D
D’
A certain medical disease occurs in 1% of the population. A simple screening procedure is
available and in 8 out of 10 cases where the patient has the disease, it produces a positive
result. If the patient does not have the disease there is still a 0.05 chance that the test will
give a positive result.
Find the probability that a randomly selected individual:
(b)
gives a positive result in the test.
= 0.008 + 0.0495 = 0.0575
0.01
0.99
0.8
P
DP
0.2
P’
DP’
0.05
P
D’P
0.95
P’
D’P’
= 0.008
D
D’
= 0.0495
A certain medical disease occurs in 1% of the population. A simple screening procedure is
available and in 8 out of 10 cases where the patient has the disease, it produces a positive
result. If the patient does not have the disease there is still a 0.05 chance that the test will
give a positive result.
Ann has taken the test and her result is positive.
(c)
Find the probability that she has the disease.
0.008
0.0575
0.01
0.99
= 0.139 (3dp)
0.8
P
DP
0.2
P’
DP’
0.05
P
D’P
0.95
P’
D’P’
= 0.008
D
D’
= 0.0495