Question

A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4 , respectively.

Answer 1

Solution. Consider the following events:
CA = “A solved problem correctly”
CB = “B solved problem correctly”
S = “A and B obtain the same answer”
C = “Their answer is correct”
E = “A and B made common error”.
We will assume that the events CA, CB, and E independent, that is
P(CACBE) = P(CA) ∗ P(CB) ∗ P(E),
P(CACB) = P(CA) ∗ P(CB), P(CAE) = P(CA) ∗ P(E), P(CBE) = P(CB) ∗ P(E).
We should find the conditional probability P(C|S). By definition where SC is the intersection
of events S and C:
SC = “A and B obtain correct answer”
By assumption we have that
P(CA) = 1
3
, P(CB) = 1
4
, P(E) = 1
20
.
We will express the probabilities P(SC) and P(S) via P(CA), P(CB), and P(CA CB).
Notice that events SC and CACB coincide:
SC = CACB = “A and B obtain correct answer”,
so
P(SC) = P(CACB) = P(CA) ∗ P(CB) = 1
3
∗
1
4
=
1
12
.
Now compute P(S). Notice that S happens if either of the following two events hold:
CACB = “A and B made no errors’
CACBE = “A and B made error and this error is the same”.
These events are mutually exclusive, that is
P(S) = P(CACB) + P(F).
Moreover,
P(F) = P(CACBE).
Since events CA, CB and E are independent, we obtain that
P(F) = P(CACBE) == P(CA) ∗ P(CB) ∗ P(E)
=

1 −
1
3

∗

1 −
1
4

∗
1
20
=
2
3
∗
3
4
∗
1
20
=
1
40
,
so
P(S) = P(CACB) + P(F) = 1
3
∗
1
4
+
1
40
=
1
12
+
1
40
=
13
120
.
Therefore
P(C|S) = P(SC)
P(S)
=
P(CACB)
P(CACB) + P(F)
=
1
12
13
120
=
10
13
.
Answer. D) 10
13 .

Answer 2

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