One coin in a collection of 13 is rigged to have a probability of showing heads equal to 0.6. The rest are fair (probability 0.5). A coin is chosen at random from the lot and tossed 2 times. It turns up heads 2 times in a row. What is the probability that it is the rigged coin given that it showed heads in 2 out of 2 tosses? ⠀ (for best results enter an exact answer or an approximation with at least 4 decimal places) Hints

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Chapter1: Combinatorial Analysis
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Need help calculating. probabilities. I don't understand what it is asking.

Since there is one rigged coin among the 13 available,
Compute
and
P(Pick rigged coin)
P(Pick fair coin)
=
13
12
13
P(Get 2 heads | Pick Rigged coin)
P(Get 2 heads | Pick Rigged coin)
using restricted sample spaces. For each of these conditional probabilities, you know what coin you are using, so you
know the probability of getting heads with that coin.
Transcribed Image Text:Since there is one rigged coin among the 13 available, Compute and P(Pick rigged coin) P(Pick fair coin) = 13 12 13 P(Get 2 heads | Pick Rigged coin) P(Get 2 heads | Pick Rigged coin) using restricted sample spaces. For each of these conditional probabilities, you know what coin you are using, so you know the probability of getting heads with that coin.
One coin in a collection of 13 is rigged to have a probability of showing heads equal to 0.6. The rest are fair
(probability 0.5). A coin is chosen at random from the lot and tossed 2 times. It turns up heads 2 times in a row. What
is the probability that it is the rigged coin given that it showed heads in 2 out of 2 tosses?
(for best results enter an exact answer or an approximation with at least 4 decimal places)
Hints
Recall:
P(F|E)
so by solving for the intersection's probability,
=
=
P(E and F)
P(F|E) · P(E).
This is a Bayes (inverse) probability problem with two experiments: choosing a coin (fair or rigged) and counting the
number of heads in 2 tosses (0 to 2 heads). You want
P(Pick Rigged coin get 2 heads) :
=
P(E and F)
P(E)
=
9
P(Pick Rigged coin AND get 2 heads)
P(Get 2 heads)
where (using cases)
P(Get 2 heads):
and (using the second formula in this hint)
.
P(Pick Rigged coin AND get 2 heads) = P(Get 2 heads | Pick Rigged coin) · P(Pick rigged coin),
P(Pick Fair coin AND get 2 heads) = P(Get 2 heads | Pick Rigged coin) · P(Pick fair coin).
Since there is one rigged coin among the 13 available,
P(Pick Rigged coin AND get 2 heads) + P(Pick Fair coin AND get 2 heads)
Transcribed Image Text:One coin in a collection of 13 is rigged to have a probability of showing heads equal to 0.6. The rest are fair (probability 0.5). A coin is chosen at random from the lot and tossed 2 times. It turns up heads 2 times in a row. What is the probability that it is the rigged coin given that it showed heads in 2 out of 2 tosses? (for best results enter an exact answer or an approximation with at least 4 decimal places) Hints Recall: P(F|E) so by solving for the intersection's probability, = = P(E and F) P(F|E) · P(E). This is a Bayes (inverse) probability problem with two experiments: choosing a coin (fair or rigged) and counting the number of heads in 2 tosses (0 to 2 heads). You want P(Pick Rigged coin get 2 heads) : = P(E and F) P(E) = 9 P(Pick Rigged coin AND get 2 heads) P(Get 2 heads) where (using cases) P(Get 2 heads): and (using the second formula in this hint) . P(Pick Rigged coin AND get 2 heads) = P(Get 2 heads | Pick Rigged coin) · P(Pick rigged coin), P(Pick Fair coin AND get 2 heads) = P(Get 2 heads | Pick Rigged coin) · P(Pick fair coin). Since there is one rigged coin among the 13 available, P(Pick Rigged coin AND get 2 heads) + P(Pick Fair coin AND get 2 heads)
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