uestion 1156743
A box contains three coins, two of them fair and one two-headed. A coin is
selected at random and tossed. If heads appear the coin is tossed again; if
tails appear then another coin is selected from the two remaining coins and
tossed.
(i) find the probability that heads appear twice.
(ii) if the same coin is tossed twice, find the probability that it is the
two-headed coin.
(iii) find the probability that tails appear twice.
In each case we select a coin, toss a coin, select a coin, then toss a coin.
Sometimes it is certain (probability = 1) which coin we select, sometimes not.
Sometimes it's certain that we'll get a head, (probability = 1), sometimes not.
There are six cases. I can't make a tree diagram here, but you can and should.
Here are the six cases:
Select(Prob.)|Toss(Prob.) |Select(Prob.)| Toss(Prob.)| Probability of possibility
-------------------------------------------------------------------------------------
1. Fair(2/3) | Head(1/2) | Fair(1) | Head(1/2) |(2/3)(1/2)(1)(1/2) = 1/6
2. " | " | " | Tail(1/2) |(2/3)(1/2)(1)(1/2) = 1/6
3. " | Tail(1/2) | Fair(1/2) | Head(1/2) |(2/3)(1/2)(1/2)(1/2) = 1/12
4. " | " | " | Tail(1/2) |(2/3)(1/2)(1/2)(1/2) = 1/12
5. " | " | 2headed(1/2)| Head(1) |(2/3)(1/2)(1/2)(1) = 1/6
6. 2headed(1/3) | Head(1) | 2headed(1) | Head(1) |(1/3)(1)(1)(1) = 1/3
Notice that the sum of the probabilities of all six cases is 1, i.e.,
1/6+1/6+1/12+1/12+1/6+1/3 = 2/12+2/12+1/12+1/12+2/12+4/12 = 12/12 = 1.
(i) find the probability that heads appear twice.
P(case 1 OR case 6) = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
(ii) if the same coin is tossed twice, find the probability that it
is the two-headed coin.
P(2-headed|tossed twice) = P(case 6|case 1 or case 6) =
P(case 6 AND case 1 or 6) P(case 6) 1/3
--------------------------- = ---------------- = ----- = (1/3)∙(2/1) = 2/3
P(case 1 OR case 6) 1/2 [from (i)] 1/2
(iii) find the probability that tails appear twice.
P(case 4) = 1/12
Edwin
