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Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin. Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’

if any other number comes

is its mean to say other than 3 and 6???

yes, other than multiple of 3 & 6 i.e. 1,2,4 and 5.
0 probability?
How??

Sample space would be {(3,1),(3,2),(3,3),(3,4),..........(6,6),(1,H),(1,T),(2,H),(2,T).....(5,H),(5,T)}
Events  : Coin shows tail {(1,T), (2,T), (4,T), (5,T)}
Atleast one die shows 3 :{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)}
Thus interesection of both is 0
Thus [probability of the event ‘the coin shows a tail / at least one die shows a 3]=0

Yes.. Correct
I didn't find till how many we to roll die.

Secondly if suppose output be like (3,1) means firsr die show 3 second time it shows 1 then we will toss a coin we have 1/2 probability of each Head and tail. How things going to zero ?
The condition of throwing a die or tossing a coin is only based on first throw not on the succesive one`s!
Got it?
May be you are right but there is no hint in question that those conditions of "throwing another die or tossing a coin" is only followed by first roll, can you please explain why you deduced like this ?

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