Welcome! We have this coin tossing problem here.
A coin is tossed twice. And we want to answer the questions listed here.
I will first answer these questions intuitively and then show how to use probability concepts
to answer them. If we toss a coin twice the possible outcomes
are Head-Head, Head-Tail, Tail-Head, and Tail-Tail. Part a). What is the probability that a head
would result on the first toss? Since there are only 2 equally likely outcomes
(Head and Tail) when a coin is tossed, the probability of getting a head on the first
toss is simply half or 1 over 2.
Part b). What is the probability that a tail would result on the second toss given that
the first toss resulted in a head? If the first toss resulted in a Head, there
are only 2 ways this could happen: HH or HT Since the second toss resulted in a tail in
only of these outcome (HT), then the probability is 1 over 2.
Part c). What is the probability that two tails would result?
Since two tails occurred only once in the 4 outcomes listed above, the probability of
obtaining 2 tails is 1 over 4.
Part d). What is the probability that a tail would result on the first toss and a head
on the second toss. Again, Tail-Head occurs only once in the 4
outcomes listed above, thus probability is 1 over 4.
Part e). What is the probability that a tail would result on the first toss and a head
on the second toss or a head on the first and a tail on the second toss?
We’re simply required to calculate the probability of Tail-Head or Head-Tail.
These represent two of the four possible outcomes. Therefore the probability is 2 over 4 which
equals a half. Part f). What is the probability that at least
one head would occur on the two tosses. There is at least one Head in 3 of the 4 outcomes.
They are: HH, HT, and TH. Therefore the probability of obtaining at
least one head is 3 over 4. Now let’s use a tree diagram together with
a few probability notations to answer these questions.
Part a) again: What is the probability that a head would result on the first toss?
On a tree diagram, the first toss of the coin looks like this. And the probability is clearly
1 over 2 or 0.5 Now let’s extend the tree to a second toss:
So if the first toss resulted in a head, since the events are independent, that is,
the result of the second toss is not affected by the first, then the probability of Head
and Tail will remain ½ each. The same applies when the first toss was a
tail. In this case we can calculate the joint probability
of 2 heads, by multiplying the 2 probabilities. These are independent events. Head-Tail, Tail-Head,
and Tail-Tail. The results are all 1 over 4, indicating that
the outcomes are equally likely – just like we treated them earlier.
Part b). What is the probability that a tail would result on the second toss given that
the first toss resulted in a head? That is, we need to find the probability of
Tail (on the second) given Head (on the first). By formula, this is Probability of H and T
divided by the probability of H. That is, a quarter, divided by half, and that
gives 0.5
Part c). What is the probability that two tails would result?
It can be seen directly from the probability tree that the probability of Tail-Tail is
¼ or 0.25
Part d). What is the probability that a tail would result on the first toss and a head
on the second toss. This can also be seen directly on the tree
that the probability of Tail-Head is ¼ or 0.25
Part e). What is the probability that a tail would result on the first toss and a head
on the second toss or a head on the first and a tail on the second toss?
We need to calculate the probability of Tail-Head or Head-Tail.
In probability, “or” means “add”. That is we add the probability of TH and HT
which gives ¼ + ¼ or ½.
Part f). What is the probability that at least one head would occur on the two tosses.
As seen on the tree, the possibilities are Head-Head, Head-Tail, or Tail-Head.
Therefore the probability of obtaining at least one head is 1/4 +1/4+1/4 which equals
3/4 or 0.75
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