#24 – Probability (Unfair Coin)
Gloria has an unfair coin. Let \( p \) be the probability of heads, with \( \tfrac{1}{2} < p < 1 \). When the coin is flipped three times, the probability of getting exactly two heads is twice the probability of getting exactly two tails. Find the probability of tails when the coin is flipped.
Show Solution
Let \( q = 1 - p \).
\[ P(\text{2 heads}) = 3p^2q, \quad P(\text{2 tails}) = 3q^2p. \] Given \( 3p^2q = 2(3q^2p) \Rightarrow p^2q = 2q^2p. \)
Divide by \( pq \) (nonzero) → \( p = 2q \).
Then \(q = 1-2q \)
Hence \( q = = \boxed{\tfrac{1}{3}}. \)
\[ P(\text{2 heads}) = 3p^2q, \quad P(\text{2 tails}) = 3q^2p. \] Given \( 3p^2q = 2(3q^2p) \Rightarrow p^2q = 2q^2p. \)
Divide by \( pq \) (nonzero) → \( p = 2q \).
Then \(q = 1-2q \)
Hence \( q = = \boxed{\tfrac{1}{3}}. \)
#26 – Expected Value (Sum of Cubes)
Johnny rolls a fair six-sided die six times and sums the cubes of all results. What is the expected value of this total?
Show Solution
Let \( X_i \) be the result of the \( i^\text{th} \) roll.
We want \( E[T] = E[X_1^3 + X_2^3 + \cdots + X_6^3] \).
By linearity of expectation: \[ E[T] = 6 \, E[X^3]. \]
By linearity of expectation: \[ E[T] = 6 \, E[X^3]. \]
\[
\displaystyle E[X^3]
= \frac{1^3+2^3+3^3+4^3+5^3+6^3}{6}
\]
\[ = \frac{(1+2+3+4+5+6)^2}{6} = \frac{21^2}{6} = \frac{441}{6} = 73.5. \] Therefore E[T] = 6 times 73.5 = 441 -- the answer.
\[ = \frac{(1+2+3+4+5+6)^2}{6} = \frac{21^2}{6} = \frac{441}{6} = 73.5. \] Therefore E[T] = 6 times 73.5 = 441 -- the answer.
(This uses the identity \( \sum_{k=1}^{n} k^3 = \big(\sum_{k=1}^{n} k\big)^2 \).)
#30 – Powers and Estimation
Find the integer \( n \) such that \[ n^7 = 44{,}231{,}334{,}895{,}529. \]
Show Solution
Since \(80^7 \approx 2.1\times10^{13}\) and \(90^7 \approx 4.78\times10^{13}\), \(n\) is between 80 and 90.
The last digit of \(n^7\) matches the last digit of \(n\), which is 9.
Modulo-9 check also gives \(n \equiv 8 \pmod9\), so the only candidate is \(n=89\).
Indeed, \(89^7 = 44{,}231{,}334{,}895{,}529\). Thus \(n = \boxed{89}.\)
Indeed, \(89^7 = 44{,}231{,}334{,}895{,}529\). Thus \(n = \boxed{89}.\)
