Two persons A and B toss a die one after another. The person who throws 6 wins . If A starts then the probability of his winning is
a)1/2
b)5/11
c)6/11
d)10/11
Answers
Answer:the correct answer will be: 1/6 which is none of the above
Step-by-step explanation:
total no. of outcomes is 6
No. of favourable outcomes is 1
(p) of getting a six should be:
no. of favourable outcomes/
total no. of outcomes
i.e 1/6
Answer:
Option(C)
Step-by-step explanation:
The person who throws 6 wins .
Probability of getting 6 = 1/6.
Probability of not getting 6 = 1 - 1/6 = 5/6.
(1) 1st die thrown by A :
Probability of A wins = (1/6)
(2) 2nd die thrown by B :
Probability of B wins = (5/6) * (1/6).
(3) 3rd die thrown by A:
Probability of A wins = (5/6) * (5/6) * (1/6)
(4) 4th die thrown by B:
Probability of B wins = (5/6) * (5/6) * (5/6) * (1/6)
(5) 5th die thrown by A:
Probability of A wins = (5/6) * (5/6) * (5/6) * (5/6) * (1/6)
Thus,
Probability of A wins will be,
=> (1/6) + (5/6) * (5/6) * (1/6) + (5/6) * (5/6) * (5/6) * (5/6) * (1/6) + ...
=> (1/6) + (5/6)^2 * (1/6) + (5/6)^4 * (1/6) + .....
Clearly, Sum is in GP with,
First term = 1/6
Common ratio = (5/6)^2
Hence, GP = a/(1 - r)^2
=> (1/6)/(1 - 5/6)^2
=> 1/6 * 36/11
=> 6/11
Therefore, probability of A winning = 6/11
Hope it helps!