Question

When a coin is tossed once find the probability of the following (1) a head (2) no head (3) a tail (4) both tails

Answer 1

Answer:

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)= TotalnumberofpossibleoutcomesNumberofoutcomesfavourabletoA​=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)​=81​

(ii) Let C be the event of the occurrence of 2 heads Accordingly C=HHT,HTH,THH

∴P(C)=n(S)n(C)​=83​

(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)​=84​=21​

(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)​=87​

(v) Let F be the event of the occurrence of no head

Accordingly F=TTT

∴P(F)=n(S)n(F)​=81​

(vi) Let G be the event of the occurrence of 3 tails

Accordingly G=TTT

∴P(G)=n(S)n(G)​=81​

(vii) Let H be the event of the occurrence of exactly 2 tails

Accordingly H=HTT,THT,TTH

∴P(H)=n(S)n(H)​=83​

(viii) Let I be the event of the occurrence of no tail

Accordingly I=HHH

∴P(I)=n(S)n(I)​=81​

(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)​=87​

Step-by-step explanation:

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