# Thread: How can i work out this probability question?

1. ## How can i work out this probability question?

Hi
Suppose i have 3 coins in a bag:two are ordinary coins,one has head on both sides.

i) i take a coin out of my bag at random and toss it.what is the probability it comes down heads?

ii)Suppose i take a coin out of my bag randomly,toss it,and it comes down heads.What is the probability that the other side is a head?

iii)What is the probability that if i toss the same coin a second time i will get another head?

Thanks i really appreciate if someone could give me an explanation of how to get there

2. Hello, math8553!

Suppose i have 3 coins in a bag: two ordinary coins, one with two heads.
I take a coin out of my bag at random.
$N$ = normal coin, . $D$ = double-head coin.

Then: . $P(N) \:=\: \tfrac{2}{3},\;\;P(D) \:=\:\tfrac{1}{3}$

a) I toss the coin.
What is the probability it comes up Heads?
There are two possibilites:

[1] The $N$-coin is chosen and it comes up Heads: . $P(N \wedge \text{Heads}) \:=\:\tfrac{2}{3}\cdot\tfrac{1}{2} \:=\:\tfrac{1}{3}$

[2] The $D$-coin is chosen and it comes up Heads: . $P(D \wedge \text{Heads}) \:=\:\tfrac{1}{3}\cdot1 \:=\:\tfrac{1}{3}$

Therefore: . $P(\text{Heads}) \;=\;\tfrac{1}{3} + \tfrac{1}{3} \;=\;\boxed{\frac{2}{3}}$

b) Suppose I take a coin randomly, toss it, and it comes up Heads.
What is the probability that the other side is a Head?
Given that coin turned up Heads, what is the probability that it is the $D$-coin?

This is Conditional Probability which requires Bayes' Theorem: . $P(A\,|\,B) \;=\;\frac{P \wedge B)}{P(B)}$

We want: . $P(D \,|\,\text{Head}) \;=\;\frac{P(D \wedge \text{Head})}{P(\text{Head})} \;=\;\frac{\frac{1}{3}}{\frac{2}{3}} \;=\;\boxed{\frac{1}{2}}$

c) )What is the probability that if i toss the same coin a second time i will get another head?

We want: . $P(\text{2nd is Head}\,|\,\text{1st is Head}) \;=\;\frac{P(\text{2nd is Head}\,\wedge\,\text{1st is Head})}{P(\text{1st is Head})}\;=\;\frac{P(HH)}{P(\text{1st is Head})}$

There are two ways to get $HH\!:$

[1] $N$-coin and two Heads: . $P(N\,\wedge\,HH) \;=\;\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{2} \;=\;\frac{1}{6}$

[2] $D$-coin and two Heads: . $P(D\,\wedge\,HH) \;=\;\frac{1}{3}\cdot1\cdot1 \;=\;\frac{1}{3}$

. . Hence: . $P(HH) \:=\:\frac{1}{6} + \frac{1}{3} \:=\:\frac{1}{2}$

Therefore: . $P(\text{2nd is Head}\,|\text{1st is Head}) \;=\;\frac{\frac{1}{2}}{\frac{2}{3}} \;=\;\boxed{\frac{3}{4}}$