## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.4 Fundamentals of Probability - Exercise Set 11.4 - Page 715: 23

#### Answer

$\frac{1}{2}$

#### Work Step by Step

P(E) = $\frac{no.\ of\ outcomes\ in\ favor}{Total\ no.\ of\ outcomes}$ Given that H is heads and T is tails, the set of equally likely outcomes is: S = {HH, HT, TH, TT}. E: The same outcome on each toss Favorable outcomes= {HH, TT} P(E) = $\frac{2}{4}$ = $\frac{1}{2}$

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