PROBABILITY

2)

probability that A wins in the first throw( as A begins 1st) = 1/6. ( as he has to choose 6 from 1,2,3,4,5,6... which can be done in one way.)
probability that A wins in the secnd attempt =(5/6)*(5/6)*(1/6).
( as it means that both A and B have failed in their 1st attempts.. this probability is 5/6 for each... as both didnt thrown 6.... and in the second attempt A succeeds with the probability 1/6)
similarly in the third attempt A wins with the probability = (5/6)*(5/6)*(5/6)*(5/6)*(1/6).
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in this way it continues as a infinite geometric progression.

so probablity = 1/6 +25/6³ + 625/6⁵............................∞
= 1/6 ( 1 + 25/36 + (25/36)² + (25/36)³.................∞)
= 1/6 * 1/(1 -25/36)
= 6/11.

thats correct...joydoot deserves a pinked post

1)Two cards are drawn at random from a well shuffled pack of 52 cards & thrown away.Find the probability of drawing an ace from the remaining 50 cards.

CASE I : two cards thrown are aces (prob. of this = 2/52 = 1/26)
in this case prob of drawing an ace = 2/50 = 1/25
So prob of this case = 1/26 * 1/25

CASE II : out of two cards. one is ace (prob of this = (4C1)*(48C1)/(52C2) )

in this case prob of drawing an ace = 3/50

so prob of this case = (4*48*2)/(52*51) * 3/50

CASE III: none of the thrown cards are aces (prob of this = (48C2)/(52C2) )

in this case probability of drawing an ace = 4/50

so prob of this case = (48*47)/(52*51)

so total prob = (48*47*4)/(52*51*50)

So, total probability = sum of prob. obtained