Both Rajesh and Ramesh have one $5, one $10 and one $20 coin each. They play a game in which each selects a coin randomly from their existing coins without the knowledge of other. If the sum of the coins is an odd amount, then Rajesh wins Ramesh’s coin and if it is an even amount then Ramesh wins Rajesh’s coin. Find the probability that Rajesh wins one odd and one even coin in the first two games
Answers
probability that Rajesh wins one odd and one even coin in the first two games = 7/36
Step-by-step explanation:
Both Rajesh and Ramesh have one $5, one $10 and one $20 coin each
Each one can select coin in 3 ways
Hence total combination = 3 * 3 = 9
5 ,5
5 , 10 Rajesh wins Even Coin
5 , 20 Rajesh wins Even coin
10 , 5 Rajesh Wins Odd Coin
10 , 10
10 , 20
20 , 5 Rajesh wins odd Coin
20 , 10
20 , 20
Case 1 : Rajesh first win odd coin
Probability that Rajesh wins odd coin = 2/9
if Rajesh wins odd coin then
Rajesh have 5 , 5 , 10 , 20 & ramesh have 10 , 20
total Combination = 8
Rajesh wins in 4 combination then Even coin
= 4/8 = 1/2
in Case 1 Probability = (2/9)(1/2) = 1/9
Case 2 : Rajesh first win Even coin
Has further two cases win 10 or 20
Probability of both = 1/9
case a : Rajesh won 10
Rajesh have 5 , 10 , 10 , 20 Ramesh has 5 & 20
Rajesh win odd coin in 3 cases out of 8
Same will have happens if he win 20 First
So probability = 2 * (1/9)(3/8) = 1/12
Total Probability = 1/9 + 1/12
= (4 + 3)/36
= 7/36
probability that Rajesh wins one odd and one even coin in the first two games = 7/36
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