Example 17 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Bayes theorem
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Misc 3
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Example 17 Important You are here
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Misc 12
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Bayes theorem
Example 17 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 and one silver coin. A person chooses a box at coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?Let B1 : Selecting Box 1 having two gold coins B2 : Selecting Box 2 having two silver coins B3 : Selecting Box 3 having one gold and one silver coin 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(B1"|"G) P(B1"|"G) = (๐(๐ต_1 ).๐(๐บ|๐ต_1))/(๐(๐ต_1 ).๐(๐บ|๐ต_1)+๐(๐ต_2 ).๐(๐บ|๐ต_2)+๐(๐ต_3 ).๐(๐บ|๐ต_3)) "P(B1)" = Probability of selecting Box 1 = ๐/๐ ๐ท(๐ฎ"|B1") = Probability that second coin is of gold in Box 1 = ๐ "P(B2)" = Probability of selecting Box 2 = ๐/๐ ๐ท(๐ฎ"|B2")= Probability that second coin is of gold in Box 2 = ๐ "P(B3)" = Probability of selecting Box 3 = ๐/๐ ๐ท(๐ฎ"|B3") = Probability that second coin is of gold in Box 3 = ๐/๐ Putting values in formula, ๐("B1|" ๐บ) = (1/3 ร 1)/( 1/3 ร 1 + 1/3 ร 0 + 1/3 ร 1/2 ) = (1/3 ร 1)/( 1/3 ร [1 + 0 + 1/2] ) = 1/( 1 + 1/2 ) = 1/( 3/2 ) = ๐/๐ Therefore, required probability is 2/3