Perms and coms (1 Viewer)

senarillosk · Nov 17, 2015

A biased coin is weighted so that
heads comes up twice as often
as tails. Find the probability of
tossing a tail.

The answer is 1/3

Can someone explain me this? Thank you.

leehuan · Nov 17, 2015

This isn't too difficult.

The probability of tossing a head or a tail must add up to one.
P(H) + P(T) = 1
But we know that heads comes up twice as likely as tails, so P(H) = 2P(T)
So 2P(T) = P(T) = 1
3P(T)=1
P(T)=1/3

InteGrand · Nov 17, 2015

senarillosk said:
A biased coin is weighted so that
heads comes up twice as often
as tails. Find the probability of
tossing a tail.

The answer is 1/3

Can someone explain me this? Thank you.

$Let the probability of a tails be $p_T$ and the probability of a heads be $p_H$. We must have $p_H + p_T = 1$, as the probabilities of all the possibilities of a coin toss (heads or tails) must sum to 1. Also, since heads comes up twice as often as tails, we have $p_H = 2p_T$ (i.e. the probability of a heads is twice that of a tails). Since $p_H = 1-p_T$, we sub. this in to the last equation to find that $1-p_T = 2p_T \Rightarrow 3p_T = 1\Rightarrow p_T = \frac{1}{3}$, as the answer says.$

Edit: done above

leehuan · Nov 17, 2015

InteGrand said:
Edit: done above

If I scrolled down further enough and saw that you were in this thread I would've left it alone lol sorry

InteGrand · Nov 17, 2015

leehuan said:
If I scrolled down further enough and saw that you were in this thread I would've left it alone lol sorry

Lol, no need to be sorry.

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Perms and coms (1 Viewer)

senarillosk

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leehuan

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InteGrand

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leehuan

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InteGrand

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