IIT JEE , AIEEE Forum - Mathematics , Algebra - probability

raman_shadow

a coin is tossed once. if tail occurs then it is tossed again repeatedly until a head shows up. after a head a dice is thrown. if a six comes up then it is rolled again until any other number comes. a person wins if he gets 1. find his probability of winning.

chimanshu_007

is the ans 5/6

master_purav

I think the ans is 1/12

master_purav

Probability of getting a head is 1/2 and that of getting 1 on dice is 1/6.

So the probability of getting both of them together is 1/2*1/6=1/12.

I m not sure...

ramyadiamond

i dont think thats right bcoz u also have to consider those cases in which he gets tails before getting a head, or gets some other no. than 6 in his attempts.

anu85

i think the answer 1/24

as the prob of getting a tail is 1/2

the prob of getting a head is 1/2

the prob of getting a six is 1/6

nd the prob of getting any no then is 1

so finally, 1/2 *1/2*1/6 =1/24

master_purav

anu85

but how can you get a head and a tail at the same time...

taruntanuj007

i think the problem needs to be approached via Baye Theorum!!!

If heads then die is cast else not then if 6 then repeatedle die is cast so we have events in succesion and the probability is not so eas to calculte using only PC well i'll try l8r

raman_shadow

hey guys i have found the answer. u have the following cases.
P(h1 h61 h661 ...)
P(th1 tth1 ttth1...)
P(th61 tth61 ttth61...)
P(th661 tth661 ttth661...)
P(th6661 tth6661 ttth6661...)
..............
..............

all cases can be solved by a/(1 - r)
the answers come as
1/10 + 1/12 + 1/72 + 1/ 432 + 1/2592 + ......
let the first 2 cases remain as they are and solve the rest again by the infinite gp formula a=1/72 r=1/6
it gives 1/60
so answer is
1/10 + 1/12 + 1/60 = 1/5

raman_shadow

hey guys a simple approach also came to my mind.
in a dice u have 6 outcomes. number 6 may not be counted as an outcome in this case because it repeats the throw.
so outcomes left = 5
favourable = 1 (u need number 1)
probability = 1/5

avinash.sharma

Let one more example: If a coin is tossed and if head comes then a dice is thrown if there is 6 on dice then a ball is drawn from the urn which contain 5 white & 4 black balls if white ball is drawn that there is a win, if there is tell then coin is tossed again, if there is non six on dice than dice is thrown again, find the probability of win.

Solution : H= Probability of head (1/2), T = Probability of Tail (1/2) , 6= Probability of getting 6 on dice (1/6), n(6)=Probability of other number on dice (5/6), W= prob. To get white ball (5/9) and n(w)= probability of getting black ball (4/9).

The required cases are :

1. H,6,W , TH6W, TTH6W,TTTH6W ......

2. H n(6) 6 W , H n(6) n(6) 6 W, H n(6) n(6) n(6) 6 W,......

3. T H n(6) 6 W, T H n(6) n(6) 6 W, T H n(6) n(6) n(6) 6 W

4....

Same way more cases. all cases can be solved by GP a/(1 - r).

now answers received from cases are again in GP which can be solved by a/(1-r)

5/54 + 25/108+25/216+.........

=5/54 + (a GP)

=5/54 + (25/108)/(1/2) = 5/54 + 25/54 = 30/54 = 5/9 Which is same as only probability of drawing a white ball from the urn.

So what is the moral of the story? It does not depend by which ways we enter into casino winning condition only depends on the performance (probability) in the casino.

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