if three coins are tossed find the probability of the following event.
event a getting at least one head and event b to get a head on third coin
Answers
Answered by
2
Answer:
Correct option is
A
Q,R
B
Q,P
n(S)=2×2×2=8
n(P)=HHT,HTH,THH,HHH=4,
n(Q)=TTT=1
n(R)=THT,HHH,HHT,THH=4
Q and R and P and Q are nutually exclusive as they have nothing in their intersection.
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Answered by
2
Answer:
A. 0.875, B. 1/2
Step-by-step explanation:
Here, Event A = Getting at least one head.
So, Event A' = Getting no head.
Hence, P(A) = 1 - P(A') = 1 - (1/2 X 1/2 X 1/2) = 7/8 = 0.875 (apx)
It implies that the event is almost certain.
Again, Event B = Getting a head on third coin
= getting a head on a coin (since here first two coins do not affect the result)
= 1/2
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