Maharashtra State BoardHSC Science (General) 11th
Advertisement Remove all ads

Solve the following: Given three identical boxes, I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold - Mathematics and Statistics

Sum

Solve the following:

Given three identical boxes, I, II, and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Advertisement Remove all ads

Solution

Let event B1: Selecting box I have two gold coins,

event B2: Selecting box II having two silver coins,

event B3: Selecting box III having one silver and one gold coin,

event G: Coin is gold.

P(B1) = P(B2) = P(B3) = `1/3`

`"P"("G"/"B"_1) = 1, "P"("G"/"B"_2) = 0, "P"("G"/"B"_3) = 1/2`

P(G) = `"P"("B"_1) "P"("G"/"B"_1) + "P"("B"_2) "P"("G"/"B"_2) + "P"("B"_3) "P"("G"/"B"_3)`

= `1/3[1 + 0 + 1/2]`

= `1/3(3/2)`

= `1/2`

To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from box I.

∴ Required probability = `"P"("B"_1/"G")`

By Bayes’ theorem,

`"P"("B"_1/"G") = ("P"("B"_1)"P"("G"/"B"_1))/("P"("G"))`

= `(1/3(1))/(1/2)`

= `2/3`

Concept: Baye'S Theorem
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 9 Probability
Miscellaneous Exercise 9 | Q II. (21) | Page 215
Advertisement Remove all ads

Video TutorialsVIEW ALL [2]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×