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Rudolph flips two fair coins, each with a half probability of getting heads.

What is the probability he flips two tails in two tosses?

So I was thinking $0.5 + 0.5 = 1$, but I got it wrong.

I guess it doesn't make sense, because he could flip heads and then flip tails, or flip tails then flip heads, or flip two heads.

How do I find the probability? Do I subtract, or divide?

SAT Math.

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  • $\begingroup$ If he had three coins, would the probability be $1.5$? $\endgroup$ – Lord Shark the Unknown Jun 14 '17 at 5:25
  • $\begingroup$ You've tried addition, subtraction, division. It is none of these, but in the same category. $\endgroup$ – Marc van Leeuwen Jun 14 '17 at 5:28
  • $\begingroup$ "I guess it doesn't make sense, because he could flip heads and then flip tails, or flip tails then flip heads, or flip two heads." yes all of these would have same probability but just because the chance of flipping a heads or tails is both 0.5 and they are independent from each other therefore getting a heads doesn't reduce or increase the chance of a tails. $\endgroup$ – Sonny Da Silva-Peters Jun 14 '17 at 5:28
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    $\begingroup$ i don't know why this is down voted... i genuinely tried to answer this question and i wrote my attempts $\endgroup$ – Saketh Malyala Jun 14 '17 at 5:31
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    $\begingroup$ You need to put more effort into understanding the problem. Asking whether you subtract, divide, etc. to get the answer is not the right way of solving this particular problem or understanding probability in general. $\endgroup$ – anomaly Jun 14 '17 at 5:33
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Denote by $T$ the event "get tail in 1 flip" and $H$ the head event. Then $\mathbb{P}(T)= \mathbb{P} (H)=1/2$. When you flip a fair coin two times, the possible outcomes are $TT$, $TH$, $HT$, $HH$ and hence the probability of get two tails in two flips, that is the number of favourable outcome over the number of all the possible outcomes, is $$\mathbb{P} (TT) = \frac{1}{4}.$$ Notice that it is equal to $\mathbb{P}(T)\cdot \mathbb{P}(T) $, why?

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    $\begingroup$ oh, because OF the chance we get the first flip tails, 1/2 of that 1/2 chance results in a second tails?? $\endgroup$ – Saketh Malyala Jun 14 '17 at 5:31
  • $\begingroup$ Because you are evaluating the probability of two events (tail in a flip and tail in another flip) that are independent (the probability of the second outcome is not influenced by the probability of the first outcome). In general, if $A$ and $B$ are independent events, $\mathbb{P} (A,B) = \mathbb{P}(A) \mathbb{P}(B)$. $\endgroup$ – mTur11 Jun 14 '17 at 5:38
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When you list possibilities for two coins, note that there are four: HH, HT, TH and TT. Each is equally likely, and one of them is the desired outcome: TT. Thus we have a probability of $\frac14$.

To obtain this from $\frac12$, the probability of obtaining tails on one flip, you would want to multiply, not add: $\frac12\times\frac12=\frac14$. In general, if the probability of one event is $p$, and the probability of a second, independent event is $q$, then the probability that both happen is $p\times q$.

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probability of head = 0.5

probability of tail = 0.5

probability is independent $\therefore$ probability of two tails $= 0.5*0.5 = {0.5}^2 = 0.25$

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    $\begingroup$ $0.25$? you mean $\endgroup$ – Saketh Malyala Jun 14 '17 at 5:29
  • $\begingroup$ yes, sorry i have edited it $\endgroup$ – Sonny Da Silva-Peters Jun 14 '17 at 5:30

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