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Difference between revisions of "1992 AHSME Problems/Problem 29"

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Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:
 
Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:
  
−
<math>Case 1</math>:(There are an odd number of heads for <math>(n-1)</math> coin flips).
+
<math>\bold{Case 1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).
 
Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.
 
Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.
  
−
<math>Case 2</math>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  
+
<math>\bold{Case 2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  
  
 
Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  
 
Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  
  
 
Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D
 
Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D
 +
 +
(Or just come up with the general formula <math>P_n=\frac{1}{2}(1+\frac{1}{3^n})</math> Simple observation really. If you really want to get mathy, prove this through induction).
  
 
~th1nq3r
 
~th1nq3r

Latest revision as of 18:29, 20 June 2021

Problem

An unfair coin has a $2/3$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?

$\text{(A) } 25\bigg(\frac{2}{3}\bigg)^{50}\quad \text{(B) } \frac{1}{2}\bigg(1-\frac{1}{3^{50}}\bigg)\quad \text{(C) } \frac{1}{2}\quad \text{(D) } \frac{1}{2}\bigg(1+\frac{1}{3^{50}}\bigg)\quad \text{(E) } \frac{2}{3}$

Solution

Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation \[P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}\] This is essentially the expansion of $\left(\frac{2}{3}+\frac{1}{3} \right)^{50}$ but without the odd power terms. To get rid of the odd power terms in $\left(\frac{2}{3}+\frac{1}{3} \right)^{50}$, we add $\left(\frac{2}{3}-\frac{1}{3} \right)^{50}$ and then divide by $2$ because the even power terms that were not canceled were expressed twice. Thus, we have \[P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)\] Or \[\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)\] which is equivalent to answer choice $\fbox{D}$.

Solution 2

Denote $P_n$ as the probability that the number of heads is even. Then we have two cases:

$\bold{Case 1}$: (There are an odd number of heads for $(n-1)$ coin flips). Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for $(n-1)$ flips?" That is precisely $1-P_{n-1}$. Now, since we want an even number of heads, what is the probability that on the $n-th$ coin flip we get a head? $\frac{2}{3}$. So we have $\frac{2}{3}(1-P_{n-1})$.

$\bold{Case 2}$: (There are an even number of heads for $(n-1)$ coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for $n$ flips will be odd. :/ So what is the probability we get a tail? $1-\frac{2}{3}=\frac{1}{3}$. Thus we have $\frac{1}{3}(P_{n-1})$.

Adding both cases, we have that $P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}$. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in $\boxed {D)}$.

Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D

(Or just come up with the general formula $P_n=\frac{1}{2}(1+\frac{1}{3^n})$ Simple observation really. If you really want to get mathy, prove this through induction).

~th1nq3r

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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