


Last updated at July 5, 2019 by Teachoo
Transcript
` Example 36 A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first. Winning the game is getting a 6 on the die P(getting 6) = 16 P(not getting six) = 1 – P(getting six) = 1 – 16 = 56 1st throw by A: A gets a six P(A wins) = 16 2nd throw by B: A does not get 6, B gets six So, P(B wins) = 56 × 16 3rd throw by A: A does not get 6, B does not get 6, A gets six P(A wins) = 56 × 56 × 16 4th throw by B: A does not get 6, B does not get 6, A does not get 6, B gets six P(B wins) = 56 × 56 × 56 × 16 5th throw by A: A does not get 6, B does not get 6, A does not get 6, B does not get 6, A gets six P(A wins) = 56 × 56 × 56 × 56 × 16 and so on So, probability that A wins is P(A wins) = 16 + 56 × 56 × 16 + 56 × 56 × 56 × 56 × 16 + ……. = 16 + 562 16 + 564 16 + ............ = 161 − 562 = 161 − 2536 = 16 36 − 2536 = 16 1136 = 611 ∴ P(A wins) = 𝟔𝟏𝟏 and P(B wins) = 1 – P(A wins) = 1 – 611 = 𝟓𝟏𝟏
Examples
Example 2
Example 3
Example 4
Example 5
Example 6 Important
Example 7 Important
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13
Example 14
Example 15
Example 16
Example 17 Important
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22
Example 23
Example 24
Example 25 Important
Example 26 Important
Example 27 Important
Example 28 Important
Example 29 Important
Example 30
Example 31 Important
Example 32 Important
Example 33
Example 34
Example 35 Important
Example 36 Important You are here
Example 37
About the Author