1thanks a lot. anyone for the other 2
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.
2)A & B play alternately with an unbiased die.He who first throws a 'six' wins.If A begins, show that his chance of winning is 611.
3)Each of the 2 identical bags contain 5 white & 5 red balls.One ball is transferred at random without notice from the 2nd bag to the first bag.Then one ball is drawn from the 1st bag.Find the probability that the ball drawn is a red one.
-
UP 0 DOWN 0 0 4
4 Answers
112)
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).
.
.
.
.
.
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.
1061)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