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Question

Die A has 4 red and 2 white faces where as die B has two red and 4 white faces. A fair coin is tossed. If head turns up, the game continues by throwing die A, if tail turns up then die B is to be used. If the first two throws resulted in red, what is the probability of getting red face at the third throw ?

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Solution

The correct option is **D** 35

Let E1 be the event that die A is used and E2 be the event that die B is used

Let E1 be the event that die A is used and E2 be the event that die B is used

Let C be the event that a red face appears in any throw.

P(E1)=12=P(E2)

P(E1)=12=P(E2)

P(C/E1)=4C16C1=23,

P(C/E2)=2C16C1=13

P(C)=P(E1)P(C/E1)+P(E2).P(C/E2)=1/2×2/3+1/2×1/3=1/2

Let D be the event that red face appears in third throw

Let E be the even that red face appears in first two throws

P(E1)=P(E2)=12

Let E be the even that red face appears in first two throws

P(E1)=P(E2)=12

P(E/E1)=23.23=(23)2,

P(D/EE1)=23

P(E/E2)=13×13=(13)2

P(D/EE2)=13

P(D/E)=P(E1)P(E/E1)P(D/EE1)+P(E2).P(E/E2)P(D/EE2)P(E1).P(E/E1)+P(E2).P(E/E2)

=1/2(2/3)2×2/3+1/2(1/3)2×1/31/2×(2/3)2+1/2×(1/3)2=35

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