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OpenStudy (anonymous):
Two identical biased coins are tossed together, and the outcome is recorded. After a large number of trials it is observed that the probability that both coins land showing tails is 0.49.
What is the probability that both coins land showing heads?
OpenStudy (anonymous):
2/4 or 1/2
OpenStudy (anonymous):
not that easy (:
OpenStudy (anonymous):
@ParthKohli can try :p
OpenStudy (anonymous):
do you know the answer @Omniscience
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OpenStudy (anonymous):
Obv.
Parth (parthkohli):
Experimental probability. I see.
OpenStudy (anonymous):
Try it (:
OpenStudy (anonymous):
why you are asking then @Omniscience
Parth (parthkohli):
An approach to this question is P(Heads) = 1 - P(Tails)
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OpenStudy (anonymous):
Yes, That was what I thought but then that is not the first step.
Parth (parthkohli):
P(Heads) = 1 - 0.49
OpenStudy (anonymous):
lol, i did the same thing but it is not right.
Parth (parthkohli):
Is this experimental or theoretical probability?
Parth (parthkohli):
I go for experimental, as it says 'biased'
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OpenStudy (anonymous):
lol, simple probability; very easy infact (:
Parth (parthkohli):
Is it somewhat like this? P(Coin will show a tails in one toss) = \(\sqrt{0.49}\)?
OpenStudy (anonymous):
Very close.
Parth (parthkohli):
Glad to hear that. Will the first step show the computation of the probability that the coin will show a 'tails' in one toss?
OpenStudy (anonymous):
Well, show the probability of throwing heads first.
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Parth (parthkohli):
1/2 is theoretical.
1 - 0.49 is experimental.
Parth (parthkohli):
Well no...
1 - sqrt0.49
OpenStudy (anonymous):
Correct, which is the probability of throwing heads.
OpenStudy (anonymous):
Then what is the probabilty that both coins land showing heads?
OpenStudy (anonymous):
Wait 1-sqrt0.49; was the probability of head.
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OpenStudy (anonymous):
one head*
Parth (parthkohli):
(1 - sqrt0.49)^2
Parth (parthkohli):
\(\Rightarrow 1^2 - 2(\sqrt{0.49})(1) + 0.49\)
OpenStudy (anonymous):
lol what
Parth (parthkohli):
\(\Rightarrow 1.49 - 2\sqrt{0.49} \)?
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OpenStudy (anonymous):
you are over thinking.
Parth (parthkohli):
Omni, that is the identity lol
\((a - b)^2 = a^2 - 2ab + b^2\)
Parth (parthkohli):
Idk... what else can I do here lol
OpenStudy (anonymous):
Don't overcomplicate the question!!
OpenStudy (anonymous):
Well, you got it right first; but you over complicated it.
The answer is \(0.09\)
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Parth (parthkohli):
But wait, that is the probability of a head in one toss.
OpenStudy (anonymous):
No that is the probability of two heads. Since once head is \(0.3\)
Parth (parthkohli):
Yep
OpenStudy (anonymous):
Do you get it?
Parth (parthkohli):
Yeah.
0.3 * 0.3
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