With the aid of a tree diagram, find the probability of obtaining ore head and tails when three coins are tossed once?

Cunnyguy

3 years ago

When three coins are tossed once the sample space is given by
S=HHH,HHT,HTH,THH,HTT,THT,TTH,TTT
∴ Accordingly n(S)=8
It is known that the probability of an event A is given by
P(A)=
Totalnumberofpossibleoutcomes
NumberofoutcomesfavourabletoA
​
=
n(S)
n(A)
​

(i) Let B be the event of the occurrence of 3 heads Accordingly B=HHH
∴P(B)=
n(S)
n(B)
​
=
8
1
​

(ii) Let C be the event of the occurrence of 2 heads Accordingly C=HHT,HTH,THH
∴P(C)=
n(S)
n(C)
​
=
8
3
​

(iii) Let D be the event of the occurrence of at least 2 heads
Accordingly D=HHH,HHT,HTH,THH
∴P(D)=
n(S)
n(D)
​
=
8
4
​
=
2
1
​

(iv) Let E be the event of the occurrence of at most 2 heads
Accordingly E=HHT,HTH,THH,HTT,THT,TTH,TTT
∴P(E)=
n(S)
n(E)
​
=
8
7
​

(v) Let F be the event of the occurrence of no head
Accordingly F=TTT
∴P(F)=
n(S)
n(F)
​
=
8
1
​

(vi) Let G be the event of the occurrence of 3 tails
Accordingly G=TTT
∴P(G)=
n(S)
n(G)
​
=
8
1
​

(vii) Let H be the event of the occurrence of exactly 2 tails
Accordingly H=HTT,THT,TTH
∴P(H)=
n(S)
n(H)
​
=
8
3
​

(viii) Let I be the event of the occurrence of no tail
Accordingly I=HHH
∴P(I)=
n(S)
n(I)
​
=
8
1
​

(ix) Let J be the event of the occurrence of at most 2 tails
Accordingly I=HHH,HHT,HTH,THH,HTT,THT,TTH
∴P(J)=
n(S)
n(J)
​
=
8
7
​