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NCERT Class XII Mathematics Chapter - - Solutions

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Determine P(E | F) in Exercises 6 to 9.
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Question : 6
Total: 101
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
(iii) E :at most two tails , F : at least one tail
Solution:  
sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }.
(i) E : ‘head on third toss’ and F : ‘heads on first two tosses’
⇒ E = {HHH, HTH, THH, TTH} and F = {HHH, HHT}
⇒ E ∩ F = {HHH}
P(E)=
4
8
=
1
2
,P(F)=
2
8
=
1
4
 and P(E∩F)=
1
8

∴P(
E
F
)=
P(E∩F)
P(F)
=
1∕8
2∕8
=
1
2

(ii) E : ‘atleast two heads’ and F : ‘atmost two heads’
⇒ E = {HHH, HHT, HTH, THH }and
F = {TTT, THT, TTH, HTT, HHT, HTH, THH}
⇒ E ∩ F = {HHT, HTH, THH}
Hence, P(E)=
4
8
=
1
2
,P(F)=
7
8
 and P(E∩F)=
3
8

∴P(E|F)=
P(E∩F)
P(F)
=
3∕8
7∕8
=
3
7

(iii) E : ‘atmost two tails’ and F : ‘atleast one tail’.
⇒ E = {HHH, HHT, HTH, HTT, THH, THT, TTH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
⇒ E ∩ F = {HHT, HTH, HTT, THH, THT, TTH}
Hence, P(E)=
7
8
,P(F)=
7
8
 and P(E∩F)=
6
8

∴P(E|F)=
P(E∩F)
P(F)
=
6∕8
7∕8
=
6
7
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