Samuel Pepys Wrote Isaac Newton Asking Which Event is More Likely to Occur | CE Board Problem in Mathematics, Surveying and Transportation Engineering

Date of Exam

November 2016

Subject

Mathematics, Surveying and Transportation Engineering

Problem
Samuel Pepys wrote Isaac Newton to ask which of three events is more likely: that a person get (a) at least 1 six when 6 dice are rolled (b) at least two sixes when 12 dice are rolled, or (c) at least 3 sixes when 18 dice are rolled. What is the answer?

A. (a) is more likely than (b) and (c)
B. (b) is more likely than (a) and (c)
C. (c) is more likely than (a) and (b)
D. (a), (b), and (c) are equally likely

Answer Key

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[ A ]

Solution

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Event (a) In rolling 6 dice:

Probability that no six will come out
$Q_0 = (5/6)^6 = 0.335$

Probability that at least 1 six will come out
$P_a = 1 - Q_0$

$P_a = 1 - 0.335 = 0.665$

Event (b) In rolling 12 dice:

Probability that no six will come out
$Q_0 = (5/6)^{12} = 0.112$

Probability that exactly 1 six will come out
$Q_1 = (1/6)^1(5/6)^{11} \times 12 = 0.269$

Probability that at least 2 sixes will come out
$P_b = 1 - Q_0 - Q_1$

$P_b = 1 - 0.112 - 0.269 = 0.619$

Event (c) In rolling 18 dice:

Probability that no six will come out
$Q_0 = (5/6)^{18} = 0.038$

Probability that exactly 1 six will come out
$Q_1 = (1/6)^1 (5/6)^{17} \times 18 = 0.135$

Probability that exactly 2 sixes will come out
$Q_2 = (1/6)^2 (5/6)^{16} \times {_{18}C_2} = 0.230$

Probability that at least 3 sixes will come out
$P_c = 1 - Q_0 - Q_1 - Q_2$

$P_c = 1 - 0.038 - 0.135 - 0.230 = 0.597$

P_a > P_b > P_c

The most likely to happen is (a) ← Answer: [ A ]

Another Solution

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The probability of getting exactly 1 six when 6 dice are rolled
$= (_6C_1)(1/6)^1(5/6)^5$

The probability of getting exactly x sixes when 6 dice are rolled
$= (_6C_x)(1/6)^x (5/6)^{6-x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, 3, \, 4, \, 5, \, 6$

The probability of getting 1 or more sixes when 6 dice are rolled
$= 1 - (_6C_0)(1/6)^0 (5/6)^{6}$

The probability of getting exactly x sixes when n dice are rolled
$= (_nC_x)(1/6)^x (5/6)^{n - x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, \ldots , \, n$

The probability of getting n or more sixes when 6n dice are rolled is

$\begin{align} \displaystyle
P & = \sum_{x = n}^{6n} ({_{6n}C_x})(1/6)^x (5/6)^{6n-x} \\
& = 1 - \sum_{x = 0}^{n - 1} ({_{6n}C_x})(1/6)^x (5/6)^{6n - x}
\end{align}$

For 6 dice, n = 1. The probability of getting 1 or more sixes is
$\begin{align} \displaystyle
P_a & = \sum_{x = 1}^{6} ({_6C_x})(1/6)^x (5/6)^{6-x} \\
& = 0.665
\end{align}$

For 12 dice, n = 2. The probability of getting 2 or more sixes is
$\begin{align} \displaystyle
P_b & = 1 - \sum_{x = 0}^{1} ({_{12}C_x})(1/6)^x (5/6)^{12 - x} \\
& = 0.619
\end{align}$

For 18 dice, n = 3. The probability of getting 3 or more sixes is
$\begin{align}
P_c & = 1 - \sum_{x = 0}^{2} ({_{18}C_x})(1/6)^x (5/6)^{18 - x} \\
& = 0.597
\end{align}$

P_a > P_b > P_c

The most likely to occur is (a) ← Answer: [ A ]