Question

three coins are flipped. the possible results are shown in the tree diagram below. what is p(2 heads, 1 tail)?

Explanation:

Step1: Count total outcomes

From the tree diagram, when flipping 3 coins, the total number of possible outcomes is \(2^3 = 8\) (or we can count from the diagram: (H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)).

Step2: Count favorable outcomes

We need the number of outcomes with 2 heads and 1 tail. Let's list them: (H,H,T), (H,T,H), (T,H,H). So there are 3 favorable outcomes? Wait, no, wait. Wait, let's check again. Wait, (H,H,T), (H,T,H), (T,H,H) – that's 3? Wait, no, wait the tree diagram: let's see the bottom row. Wait the outcomes are: (H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T). So the ones with 2 heads and 1 tail: (H,H,T), (H,T,H), (T,H,H) – that's 3? Wait no, wait 2 heads and 1 tail: the number of such combinations is given by the binomial coefficient \(\binom{3}{2}=\frac{3!}{2!(3 - 2)!}=\frac{6}{2\times1}=3\)? Wait no, wait \(\binom{3}{2}=3\)? Wait no, \(\binom{3}{2}=3\)? Wait, no, \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), so \(\binom{3}{2}=\frac{3!}{2!1!}=\frac{6}{2\times1}=3\). Wait but wait, actually, when flipping 3 coins, the number of ways to get 2 heads and 1 tail is 3? Wait no, wait no: (HHT), (HTH), (THH) – that's 3. Wait but wait, the total number of outcomes is 8. Wait but wait, maybe I made a mistake. Wait the tree diagram: let's count the number of outcomes with 2 heads and 1 tail. Let's look at the bottom row:

1. (H,H,H) – 3 heads
2. (H,H,T) – 2 heads, 1 tail
3. (H,T,H) – 2 heads, 1 tail
4. (H,T,T) – 1 head, 2 tails
5. (T,H,H) – 2 heads, 1 tail
6. (T,H,T) – 1 head, 2 tails
7. (T,T,H) – 1 head, 2 tails
8. (T,T,T) – 0 heads

So the favorable outcomes (2 heads, 1 tail) are (H,H,T), (H,T,H), (T,H,H) – that's 3? Wait no, wait that's 3? Wait no, wait 3 outcomes? Wait but the binomial probability formula: \(P(k) = \binom{n}{k}p^k(1 - p)^n\), where \(n = 3\), \(k = 2\), \(p = 0.5\) (assuming fair coins). So \(\binom{3}{2}(0.5)^2(0.5)^1=\frac{3!}{2!1!}\times\frac{1}{4}\times\frac{1}{2}=3\times\frac{1}{8}=\frac{3}{8}\). Wait but wait, in the tree diagram, how many outcomes are there? Let's count the number of leaves: 8. And the number of leaves with 2 heads and 1 tail: let's see, the third level (since 3 coins) has 8 nodes. Let's list them:

• First branch (H) then (H) then (H): (H,H,H)
• First (H), (H), (T): (H,H,T)
• First (H), (T), (H): (H,T,H)
• First (H), (T), (T): (H,T,H)? No, (H,T,T)
• First (T), (H), (H): (T,H,H)
• First (T), (H), (T): (T,H,T)
• First (T), (T), (H): (T,T,H)
• First (T), (T), (T): (T,T,T)

Wait so the ones with 2 heads and 1 tail: (H,H,T), (H,T,H), (T,H,H) – that's 3. Wait but wait, is that correct? Wait no, wait (HHT), (HTH), (THH) – that's 3. So the number of favorable outcomes is 3, total is 8. So probability is \(\frac{3}{8}\). Wait but wait, maybe I miscounted. Wait let's check again. The tree diagram:

First flip: H or T (2 options)

Second flip: for each first flip, H or T (so 2*2=4 after two flips)

Third flip: for each of the 4, H or T (so 4*2=8 after three flips)

Now, the outcomes with 2 heads and 1 tail:

• H, H, T

• H, T, H

• T, H, H

Yes, that's 3. So total outcomes: 8. So probability is \(\frac{3}{8}\).

Wait but wait, the binomial coefficient \(\binom{3}{2}\) is 3, so that's correct. So the probability is \(\frac{3}{8}\).

Answer:

\(\frac{3}{8}\)