Combinatorics and Probability 7811 −
95
100
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94
99
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93
98
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92
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≈ 0. 230.
923.We apply Bayes’ formula. LetBbe the event that the plane flying out of Eulerville
is a jet plane andA 1 , respectively,A 2 , the events that the plane flying between the two
cities is a jet, respectively, a propeller plane. Then
P(A 1 )=
2
3
,P(A 2 )=
1
3
, P (B/A 1 )=
2
7
, P (B/A 2 )=
1
7
.
Bayes formula gives
P(A 2 /B)=
P(A 2 )P (B/A 2 )
P(A 1 )P (B/A 1 )+P(A 2 )P (B/A 2 )
=
1
3 ·1
7
2
3 ·2
7 +1
3 ·1
7=
1
5
.
Thus the answer to the problem is^15.
Remark.Without the farmer seeing the jet plane flying out of Eulerville, the probability
would have been^13. What you know affects your calculation of probabilities.
924.We find instead the probabilityP (n)for no consecutive heads to appear innthrows.
We do this recursively. If the first throw is tails, which happens with probability^12 , then
the probability for no consecutive heads to appear afterward isP(n− 1 ). If the first
throw is heads, the second must be tails, and this configuration has probability^14. The
probability that no consecutive heads appear later isP(n− 2 ). We obtain the recurrence
P (n)=1
2
P(n− 1 )+1
4
P(n− 2 ),withP( 1 )=1, andP( 2 )=^34. Make this relation more homogeneous by substituting
xn= 2 nP (n). We recognize the recurrence for the Fibonacci sequencexn+ 1 =xn+xn− 1 ,
with the remark thatx 1 =F 3 andx 2 =F 4. It follows thatxn=Fn+ 2 ,P (n)=F 2 n+n^2 , and
the probability required by the problem isP (n)= 1 −Fn 2 +n^2.
(L.C. Larson,Problem-Solving Through Problems, Springer-Verlag, 1990)
925.FixN =m+n, the total amount of money, and varym. Denote byP (m)the
probability thatAwins all the money when starting withmdollars. Clearly,P( 0 )= 0
andP(N)=1. We want a recurrence relation forP (m).
Assume thatAstarts withkdollars. During the first game,Acan win, lose, or the
game can be a draw. IfAwins this game, then the probability of winning all the money
afterward isP(k+ 1 ).IfAloses, the probability of winning in the end isP(k− 1 ).
Finally, if the first game is a draw, nothing changes, so the probability ofAwinning in
the end remains equal toP(k). These three situations occur with probabilitiesp, q, r,
respectively; hence