# Math Help - Coin tossed

1. ## Coin tossed

Two coins are tossed twice each.
Find the probability that the head appears on the first toss and the same faces appear in the second toss.

2. Hello, u2_wa!

Could you restate the problem?
As given, it's a silly question . . .

Two coins are tossed twice each.
Find the probability that the head appears on the first toss
and the same faces appear in the two tosses. .
Is this right?

If there must be a Head in the first toss of the two coins
. . and the same face appears in ALL the tosses,
. . then both coins must have Heads ... both times.

Of the $2^4 = 16$ possible outcomes,
. . exactly one satisfies the condition: . $HHHH$

Answer: . $\frac{1}{16}$

3. Originally Posted by u2_wa
Two coins are tossed twice each.
Find the probability that the head appears on the first toss and the same faces appear in the two tosses.
I think I would be inclined to interpret this differently from Soroban. A head appears on at least one coin the first time they are tossed. And each coin then has, on the second toss, the same face it had on the first toss. The possibilities now are HH HH, HT HT, TH TH (first two letters are the two coins on the first toss, second two are the same coins on the secnd toss). That gives 3 possible outcomes out of 16: the probability is 3/16.

4. Originally Posted by u2_wa
Two coins are tossed twice each.
Find the probability that the head appears on the first toss and the same faces appear in the two tosses.
I think I would be inclined to interpret this differently from Soroban: two coins are tossed and at least one of them comes up heads. The two coins are tossed again and each coin has the same face it did in the first toss.
The possible outcomes are not HH HH, HT HT, TH TH where the first two letters are the faces that come up on the first toss, and the second two letters are those same coins on the second toss. There are 3 possible outcomes out of 16 so the probability is 3/16.

5. ## Coins tossed

I agree with HallsofIvy's intepretation of the question, and his final answer, but I'm afraid I don't understand his method. (Maybe it's me being a bit slow!)

I would argue as follows:

Toss 1: the probability that both coins land tails is $\frac{1}{4}$. So the probability that at least one lands heads is $\frac{3}{4}$.

Toss 2: the probability that each coin lands the same way up as it did in toss 1 is $\frac{1}{2}$. So the probability that they both do is $\frac{1}{4}$.

Therefore the probability that both of these events occur is $\frac{3}{4}\times\frac{1}{4} = \frac{3}{16}$.

6. Hello, u2_wa!

I hope I understand the problem . . .

Two coins are tossed twice each.
Find the probability that a head appears on the first toss
and the same faces appear in the second toss.
There is no formula for this . . . We must work out the possible cases.

Suppose the two coins are a Quarter and a Dime.
. . There are: . $2^4 = 16$ possible outcomes.

On the first toss, there are four possible outcomes:
. . $(Q,D) \;=\;(H,H),\:(H,T),\:(T,H),\:(T,T)$

In three of these, a Head appears on the first toss.

Now consider the second toss ... there are four outcomes.
In how many of these do we get the "same faces"?

. . $\begin{array}{c|c} \text{1st toss} & \text{2nd toss} \\ \hline (H,H) & (H,H) \\ (H,T) & (H,T) \\ (H,T) &(T,H) \\ (T,H) & (H,T)\\ (T,H) & (T,H) \end{array}\quad\hdots$ There are 5 cases.

Therefore, the probability is $\frac{5}{16}$

7. Sir I still did'nt get your answer. Let me tell you what I think about the problem:

Let A be the event head appear on the first toss. P(A)= 3/4 . Let B be the event same faces appear in second toss than B=(HH,TT). So P(B) = 1/2

Now A & B are the independent events so P(A&B) = 3/4 x 1/2 = 3/8

Is my justification correct??

8. Originally Posted by u2_wa
Sir I still did'nt get your answer. Let me tell you what I think about the problem: Let A be the event head appear on the first toss. P(A)= 3/4 . Let B be the event same faces appear in second toss than B=(HH,TT). So P(B) = 1/2. Now A & B are the independent events so P(A&B) = 3/4 x 1/2 = 3/8 Is my justification correct??
I agree with that interpretation of this problem.