Each of three identical jewelry boxes has two drawers. In each drawer of the first box there is a gold watch. In each drawer of the second box there is a silver watch. In one drawer of the third box there is a gold watch, while in the other drawer there is a silver watch. If we select a box at random, open one of the drawers, and find it to contain a silver watch, what is the probability that the other drawer has the gold watch?
Answers
Step 1:
Let E1,E2E1,E2 and E3E3 be the event that boxes I,II and III are chosen respectively.
The P(E1)=P(E2)=P(E3)=13P(E1)=P(E2)=P(E3)=13
Consider the occurrence A as the occurrence that the watch drawn in of gold.
Then P(A/E1)P(A/E1)=P(a gold watch from box I)=2222=1
P(A/E2)P(A/E2)=P(a gold watch from box II)=0
P(A/E3)P(A/E3)=P(a gold watch from box III)=12
Step 2:
The probability that the other watch in the box is of gold=the probability that gold watch is drawn from the box I
⇒P(E1/A)⇒P(E1/A)
By Baye's theorem ,we know that
P(E1/A)=P(E1)P(A/E1)P(E1).P(A/E1)+P(E2)P(A/E2)+P(E3)P(A/E3)P(E1/A)=P(E1)P(A/E1)P(E1).P(A/E1)+P(E2)P(A/E2)+P(E3)P(A/E3)
⇒1/3×11/3×1+1/3×0+1/3×1/2
⇒2/3
Hence the required probability is 2/3
Answer:
Let B
1
: selecting box 1 having two gold coins.
B
2
: selecting box 2 having two silver coins
B
3
: selecting box 3 having one gold & one silver.
G: The second coin is of Gold
We need to find the probability that the other coin in the box is also of gold, if the first coin is of gold
i.e., P(B
1 /G)P(B 1 /G)= P(B 1 )⋅P(G/B 1 )+P(B 2 )⋅P(G/B )+P(B 3 )P(G/B 3 )P(B 1 )⋅P(G/B 1 )
P(B 1 )=Probability of selecting box 1
=1/3
P(G/B 1
)= Probability that second coin is of gold in box 1=1
P(B
2
)=Probabiluty of selecting box 2
=1/3
P(G/B
2
)=Probability that second coin is of gold in box 2
=0
P(B)
3
=Probability of selecting box 3
=1/3
P(G/B
3
)= Probability that second coin is of gold in box 3
=1/2
P(B
1
/G)=
3
1
×1+0×
3
1
+
3
1
×
2
1
3
1
×1
=
3
1
+
6
1
3
1
P(B
1
/G)=
3
2
.