Probability that $6$ of $9$ coin flips yield heads given that the first flip yields tails

Question

ANSWER ONLY part (c)

Question:

When coin 1 is flipped, it lands on heads with probability   $\dfrac{4}{5}$  ; when coin $2$ is flipped it lands on heads with probability $\dfrac{7}{10}$.

(a) If coin $1$ is flipped $11$ times, find the probability that it lands on heads at least $9$ times.

(b) If one of the coins is randomly selected and flipped $9$ times, what is the probability that it lands on heads exactly $6$ times?

(c) In part (b), given that the first of these $9$ flips lands on tails, find the conditional probability that exactly $6$ of the $9$ flips land on heads.

I got the answer to part (c) as $0.19688$ but it is wrong.

My solution:

$$\left(\left.\frac45\middle/\Big(\frac45+\frac7{10}\Big)\right.\right)\times\left({^8C_5}(\frac45)^5(\frac15)^3\right)+\left(\left.\frac7{10}\middle/\Big(\frac7{10}+\frac45\Big)\right.\right)\times\left({^8C_5}(\frac7{10})^5(\frac3{10})^3\right) =019688$$

Answer 1

You gave the probability for obtaining 5 more heads among a further 8 flips given the coin showed heads on the first flip.

You wanted the probability that six among an additional eight flips show heads, given that the coin had landed on tails in the earlier flip.

The calculation method is correct. The events used were wrong.

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So the events of interest are: $A$ the event of 1 tail among 1 flip, $B$ the event of 6 heads among 8 flips, and $C_1, C_2$ the respective events of selecting the coins. $A,B$ are conditionally independent for a given coin.

This, the Law of Total probability, and Bayes' rule gives us:

$$\begin{align}\mathsf P(B\mid A) &= \mathsf P(B\mid C_1)\,\mathsf P(C_1\mid A)+\mathsf P(B\mid C_2)\,\mathsf P(C_2\mid A)\\[2ex]&=\dfrac{\mathsf P(B\mid C_1)\,\mathsf P(A\mid C_1)\,\mathsf P(C_1)+\mathsf P(B\mid C_2)\,\mathsf P(A\mid C_2)\,\mathsf P(C_2)}{\mathsf P(A\mid C_1)\,\mathsf P(C_1)+\mathsf P(A\mid C_2)\,\mathsf P(C_2)}\\[2ex]&=\dfrac{\mathsf P(B\mid C_1)\,\mathsf P(A\mid C_1)+\mathsf P(B\mid C_2)\,\mathsf P(A\mid C_2)}{\mathsf P(A\mid C_1)+\mathsf P(A\mid C_2)}\\&=\dfrac{\dbinom86\dfrac {4^6}{5^8}\cdot \dfrac 15+\dbinom86\dfrac{7^63^2}{10^8}\cdot \dfrac 3{10}}{\dfrac 15+\dfrac 3{10}}\end{align}$$

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Answer 2

Questions like these can be tricky, so let's check our reasoning at each step. Intuitively, given the event that the first of the 9 flips is tails, it's more likely that we chose coin 2 instead of coin 1. Let "1 = T" be the event that the first flip is tails, and $C_i$ be the event that we pick coin $i$. Indeed, we have $$ P(C_1 | 1 = T) = \frac{P(C_1, 1 = T)}{P(1 = T)} = \frac{P(C_1, 1 = T)}{P(C_1, 1 = T) + P(C_2, 1 = T)} = \frac{(0.5) (0.2)}{(0.5)(0.2) + (0.5)(0.3)} = 0.4.$$

Great. Now let $H_{6, 9}$ be the event that we flip exactly 6 out of 9 heads. Note that the event ($H_{6, 9}$ and (1 = T)) seems to be precisely the event $H_{6, 8}$. Let's try computing the desired quantity (trap method!):

$$ \begin{align*} P(H_{6, 9}| 1 = T) &= \frac{P(H_{6, 9}, 1 = T)}{P(1 = T)} \\ &= \frac{P(H_{6, 8})}{P(1 = T)} \\ &= \frac{P(C_1, H_{6, 8}) + P(C_2, H_{6,8})}{P(C_1, 1 = T) + P(C_2, 1 = T)} \\ &= \frac{(0.5) \cdot \binom{8}{6}(0.8)^6(0.2)^2 + (0.5) \cdot \binom{8}{6}(0.7)^6(0.3)^2}{(0.5)(0.2) + (0.5)(0.3)} \\ &\approx 1.18. \end{align*}$$

This is clearly wrong! Where did we go wrong?

• For starters, we shouldn't have $P(H_{6, 9}, 1 = T) > P(1 = T)$, since we should be able to split $P(1 = T)$ up into $P(H_{6, 9}, 1 = T) + P(not H_{6, 9}, 1 = T)$, similar to how we computed $P(C_1 | 1 = T)$.
• We also have a nagging feeling that we didn't properly utilize our intuition from our first calculation; the information that "coin 2 is more likely" doesn't really appear in our formula.

Let's try visualizing the essence of what conditional probability does. Consider the sample space of all the possible outcomes of this experiment (example of an outcome: 'Coin 1, T, T, H, H, T, H, T, T, H').

• When we condition by an event, we reduce the valid sample space. Here, when we first condition by the event "1 = T", we remove all worlds/all outcomes where the first flip was heads. Since our coins were already biased towards heads, this takes out a lot of possibilities - the total probability of "1 = T" happening from the start is (0.5)(0.2) + (0.5)(0.3) = 0.25, which seems reasonable. In other words, from the fact that we know "1 = T", we've eliminated 75% of the total sequences.
• Furthermore, among these 25% remaining sequences, 60% of them have coin 2 as the starting coin! Our original sample space had half of the sequences using coin 1, and the other half of the sequences using coin 2. But conditioning by the event '1 = T' wipes out more of the coin 1 sequences than the coin 2 sequences!

It's now clear that our numerator was the issue. We need the proportion of sequences among the ones that are left that satisfy 6 heads out of the next 8! The correct formula is

$$\frac{(0.5)(0.2) \cdot \binom{8}{6}(0.8)^6(0.2)^2 + (0.5)(0.3) \cdot \binom{8}{6}(0.7)^6(0.3)^2}{(0.5)(0.2) + (0.5)(0.3)} \approx \boxed{0.29}.$$

The issue was our reasoning that the event ($H_{6, 9}$ and (1 = T)) equaled the event $H_{6, 8}$. A brand new instance of $H_{6, 8}$ forgets that the improbable event (1 = T) happened, and lets in way too many new sequences into our numerator. The subtlety lies in the fact that $H_{6, 9}$ includes the result of 1!! Rather, we should instead have ($H_{6, 9}$-including-1 and (1 = T)) = ($H_{6, 8}$-excluding-1 and (1 = T)) = ($H_{6, 8}$ and (1 = T)).