A coin is tossed twice if the coin shows head it is tossed again
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if the coin shows head it is tossed again but it it shows a tail then a die is tossed
Sample space={HH,HT,(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}
Total number of outcomes=8
Let A be even that die shows a number greater than 4
A={(T,5),(T,6)}
B be event that first thrown of the coin results in a tail
B={(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}
We have to find the probability
Hence, the probability that the die shows a number greater than 4,if it known that the first throw of the coin results in a tail=0.33
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Answer:
If the coin is tossed twice the possible out comes as follows.
S ={(H,H),(H,T),(T,T),(T,H)}
E be the event of getting (H,H)
Therefore n=4, m=1
P(E)=m/n=1/4
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