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An unfair coin which has 0.35 probability to result head is tossed four times.

Build and represent graphically the probability distribution and the cumulative distribution of the function with random variable "X = number of time result is head".

This is what I did:

X   |   P                               |   Cumulative
0   |   0.35^0 x 0.65^4 = 0.17850625    |   0.17850625
1   |   0.35^1 x 0.65^3 = 0.09611875    |   0.274625
2   |   0.35^2 x 0.65^2 = 0.05175625    |   0.32638125
3   |   0.35^3 x 0.65^1 = 0.02786875    |   0.35425
4   |   0.35^4 x 0.65^0 = 0.01500625    |   0.36925625
                        = 0.36925625    | = 1.50301875

Of course 0.65 is tail (1-0.35). But there's something wrong.. Shouldn't the sum of probabilities give 1? They're all possible cases (X=0 is the probability to have 0 times head in 4 tosses of the coin, X=1 is the probability to have 1 time head in 4 tosses of the coin, etc)...

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Except from the two extremes, i.e. the events $\{H,H,H,H\}$ and $\{T,T,T,T\}$, the intermediate events can happen in more ways than one, and this is because your random variable does not specify the exact tosses where head or tails occur, it is concerned only with their number of appearances. So for example, the event "3 heads" can happen in 4 different ways: $\{T,H,H,H\}, \{H,T,H,H\},\{H,H,T,H\},\{H,H,H,T\}$, and each of these events has probability $(0.35)^3\cdot 0.65$. So $P(3 Heads) = 4\cdot(0.35)^3\cdot 0.65 $

I guess you can take it from there.

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  • $\begingroup$ Ok it's clear now. :) (0.178x1)+(0.0961x4)+(0.0518x6)+(0.0279x4)+(0.015x1)=1 $\endgroup$ – MultiformeIngegno Sep 25 '14 at 14:50
  • $\begingroup$ Yes, that's it. $\endgroup$ – Alecos Papadopoulos Sep 25 '14 at 14:52

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