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In a coin toss experiment, a coin has a 30% chance of landing on H and 70% chance on landing on T

We toss the coin twice

E represents event in which at least 1 of the tosses land on head

F represents event in which at least 1 of the tosses land on tail

I can prove mathematically that they are not independent, but can someone explain to be in layman's term why they aren't?

Proof that they are not independent

Pr(E) =.51 Pr(F) =.91

E $\bigcap$ F = { HT, TH} = .21 + .21 = .42 $\neq$ Pr(E) $\cdot$ Pr(F)

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  • $\begingroup$ how do i start a new line? I hit "enter" in the editor but Pr(E) =.51 Pr(F) =.91 E $\bigcap$ F = { HT, TH} = .21 + .21 = .42 $\neq$ Pr(E) $\cdot$ Pr(F) are still on the same line.... $\endgroup$ – user133466 Sep 22 '12 at 15:50
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    $\begingroup$ It is not correct to say that $E ⋂ F = \{ HT, TH\} = 0.21 + 0.21$. Rather, one should say $\Pr(E ⋂ F) = \Pr\{HT, TH\} = 0.21 + 0.21$. An event is not a number; a probability is a number. $\endgroup$ – Michael Hardy Sep 22 '12 at 18:54
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If I tell you that event E happened, does that give you any information on event F? For example, if I roll a die and toss a coin, let E={rolled a 3} and F={got Heads}. If I tell you I rolled a 3, does that give any information on whether or not you got Heads? No, so these two events are independent. Now think of your case. If I ask the probability that F happens, you'll calculate the probability based on the fact that HH, HT, or TH (or TT) happening is a possibility. If I then tell you that E did not happen, and ask you to calculate the probability of F happening, you can throw out those three possibilities. If E didn't happen, you must have gotten the string TT, so $P(F|E)=1$. By knowing something about E, we can say something about F, so the two are dependent.

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  • $\begingroup$ ok, this is a clear explanation,thank you! I'll accept the answer when the system allows $\endgroup$ – user133466 Sep 22 '12 at 15:54
  • $\begingroup$ any idea on how to make a carriage return in the editor? $\endgroup$ – user133466 Sep 22 '12 at 15:55
  • $\begingroup$ Use a double return. $\endgroup$ – Michael Dyrud Sep 22 '12 at 15:59

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