Determine P(EIF) in Exercises 6 to 9.
A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads. F:at most two heads
(ii) Eat most two tails, F: at least one tail
Answers
Answered by
0
Answer:
If a coin is tossed three times, then the sample space S is
S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
It can be seen that the sample space has 8 elements.
(i) E={HHH,HTH,THH,TTH}
F={HHH,HHT}
∴E∩F={HHH}
P(F)=
8
2
=
4
1
and P(E∩F)=
8
1
P(E∣F)=
P(F)
P(E∩F)
=
4
1
8
1
=
8
4
=
2
1
=0.50
(ii) E={HHH,HHT,HTH,THH}
F={HHT,HTH,HTT,THH,THT,TTH,TTT}
∴E∩F={HHT,HTH,THH}
Clearly, P(E∩F)=
8
3
and P(F)=
8
7
P(E∣F)=
P(F)
P(E∩F)
=
8
7
8
3
=
7
3
=0.42
(iii) E={HHH,HHT,HTT,HTH,THH,THT,TTH}
F={HHT,HTT,HTH,THH,THT,TTH,TTT}
∴E∩F={HHT,HTT,HTH,THH,THT,TTH}
P(F)=
8
7
and P(E∩F)=
8
6
Therefore, P(E∣F)=
P(F)
P(E∩F)
=
8
7
8
6
=
7
6
=0.85
Step-by-step explanation:
plz make me beainliest
Similar questions