Probability problem on Coin
Examples 1 :
We tossed a single coin on the air and Find the probability of getting head ?
Answer :
Here n(S) is Total number of Possible outcomes So,
S = { H, T }.
n(E) Total number of required outcomes. n(E)= 1{H}.`
P(E)Probability of Event.
P(E) = n(E) / n(S) = 1 / 2.
Example 2 : Two Coins are impartial way throw on air and Find the Probability of 1 head
Answer :
Here n(S) is Total number of Possible outcomes So,
S = { H H, HT, TH, T T }.
E = Event of getting at most one head.
E = {T T,HT,TH}.
P(E) = n(E) / n(S) = 3 / 4.
Example 3:
Two coins are tossed . What is the probability of the appearing of at most one head ?
Answer :
n ( S ) = 4 = { (T , T) , ( H , T ) , ( T , H ) , ( H , H ) }
probability of the appearing of at most one head = { HH , HT , TH , TT } ,
p ( E ) = 3 / 4 .
Example 4:
Two coins are tossed . What is the probability of the appearing of at most two head ?
Answer :
n ( S ) = 4 = { (T , T) , ( H , T ) , ( T , H ) , ( H , H ) }
probability of the appearing of at most two head = { HH , HT , TH , TT },
n ( E ) = 4 .
So , p ( E ) = 4 / 4 = 1 .
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