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Thread: How can i work out this probability question?

  1. #1
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    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
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  2. #2
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    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.
    $\displaystyle N$ = normal coin, . $\displaystyle D$ = double-head coin.

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


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

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

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

    Therefore: .$\displaystyle 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 $\displaystyle D$-coin?

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

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




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

    We want: .$\displaystyle 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 $\displaystyle HH\!:$

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

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

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


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

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