1for q2> all above ans incorrect...
Answer:
3/8
On an average out of 12 games of chess played by A and B, A wins 6, B wins 4 and 2 games end in a tie.
A and B play a tournament of 3 games. Calculate the probability that A and B win alternate game, no game is tied up.
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18 Answers
21my ans is coming to be 1/8.
sol-> probability(p)=1/8*4C0*(1/2)4+ 1/8*4C1*(1/2)4 + 1/8*4C2*(1/2)4 + 1/8*4C3*(1/2)4 + 1/8*4C4*(1/2)4
or,p=1/16*1/8*16
or,p=1/8
49ohh sorry i get i am wrong!!
i have included the case when only the first four coins give head twice!!
so this method is wrong!!
strifing for a right answer!!
49anyway i will do mine!!
so, no of possible outcomes
for each coin tossed, there are 2 outcomes possible, heads or tails.
so total possible outomes is 27 for 7 coins.
now for the first three coins to be heads, no of possible outcomes is, 24
now, the set can strt wid 1 2 3 4 or 5
so total favourable outcomes is 5X24
P(E)=58
1Question 2
A fair coin is tossed 7 times, then probability that atleast 3 consecutive heads occur is equal to
1did u take no tie in tournament as >
there has to be a winner in each game for SURE
i took it as >
5/6 for no tie in the game