Coin A is a fair coin and coin B is weighted so that the probability of obtaining heads is 0.6. An experiment consists of randomly choosing a [single, fixed] coin and then flipping it 10 times. Given that the coin came up heads exactly 6 times in the 10 flips, what is the probability that you selected the weighted coin?
For each coin this is a pint binomial probability, not cumulative.
For the fair coin, p=0.5, q=0.5, x= 6, n = 10, and I get P( 6 tosses ) = 0.20508 $\approx 0.205$, A = choose fair coin.
For the biased coin, p=0.6, q=0.4, x= 6, n = 10, and I get P( 6 tosses ) = 0.25082 $\approx 0.251$, B = choose biased coin
Method A. However, for P( I chose biased coin ), it reads like a conditional probability. $$ P( B|A) = \frac{P(\space A\space and\space B\space)}{P(\space B \space)}$$ But the numerator for conditional probability is P ( fair and biased) = 0. You can't chose both coins- fair and biased. So obviously, this is incorrect. Method B. So I worked it this way- $$ P( chose\space biased\space coin) = \frac{0.251}{0.251 + 0.205}\approx 0.5504$$ Is this correct? Please clarify and elaborate on the distinctions between the two methods.
Source: I'm a tutor and this came from a student's worksheet.
Clearly defining the events of interest and using proper terminology and notation is important for communicating mathematics and arriving at a clear answer. At no point do you define what you mean by "A" and "B." You don't define what your variables represent. You don't state your assumptions and the model. This sort of "computation without explanation" is perplexingly common among students and is, in my opinion, a symptom of an answer-focused approach to doing math, as opposed to process-focused.
Here is how the reasoning might proceed. Let $A$ be the event that the fair coin is selected, and $B$ be the complementary event that the biased coin is selected. Let $n = 10$ be the number of flips of whichever coin is selected. If event $A$ occurs, then $p = 0.5$ is the probability that the coin lands heads. If event $B$ occurs, then $p = 0.6$ is the probability that the coin lands heads. Let $X$ be a random variable that counts the number of heads obtained in $n = 10$ flips. Then we model the number of heads observed with a binomial distribution conditional on which coin was selected, i.e. $$\begin{align}X \mid A &\sim \operatorname{Binomial}(n = 10, p = 0.5) \\ X \mid B &\sim \operatorname{Binomial}(n = 10, p = 0.6). \end{align}\tag{1}$$
With this notation established, we are asked for $$\Pr[B \mid X = 6],$$ the posterior conditional probability that, given we observed $X = 6$ heads, that event $B$ occurred--i.e., we selected the biased coin. Now we are ready to proceed with the computation, via Bayes' rule:
$$\Pr[B \mid X = 6] = \frac{\Pr[X = 6 \mid B]\Pr[B]}{\Pr[X = 6]}. \tag{2}$$ The denominator, which is the unconditional or marginal probability of obtaining $6$ heads, is calculated from the law of total probability, in which we condition on which coin was selected:
$$\Pr[X = 6] = \Pr[X = 6 \mid A]\Pr[A] + \Pr[X = 6 \mid B]\Pr[B]. \tag{3}$$ Note that $A$ and $B$ are complementary events that encompass the entire set of outcomes for the coin selection process.
From $(1)$, we can readily calculate the conditional probabilities
$$\begin{align}\Pr[X = 6 \mid A] &= \binom{10}{6} (1/2)^{6} (1 - 1/2)^{10 - 6} = \frac{105}{512} \approx 0.205078, \\ \Pr[X = 6 \mid B] &= \binom{10}{6} (6/10)^{6} (1 - 6/10)^{10 - 6} = \frac{489888}{1953125} \approx 0.250823. \end{align}\tag{4}$$
As for $\Pr[A]$ and $\Pr[B]$, the question strongly implies but does not explicitly state that the coin selection is equiprobable--we are told the coin is chosen "at random" which suggests that $$\Pr[A] = \Pr[B] = \frac{1}{2}, \tag{5}$$ which means that either coin is equally likely to be chosen than the other. In such a case, the posterior probability simplifies to $$\Pr[B \mid X = 6] = \frac{\Pr[X = 6 \mid B]}{\Pr[X = 6 \mid A] + \Pr[X = 6 \mid B]} \approx \frac{0.250823}{0.205078 + 0.250823} \approx 0.550169.$$
The exact value is $$\Pr[B \mid X = 6] = \frac{\frac{3^6 \cdot 2^4}{5^{10}}}{\frac{1}{2^{10}} + \frac{3^6 \cdot 2^4}{5^{10}}} = \frac{3^6 \cdot 2^{14}}{5^{10} + 3^6 \cdot 2^{14}} = \frac{11943936}{21709561}.$$
Regarding your "Method A," it I have no idea what you are trying to say, because you did not define the events of interest. If I interpret $A$ and $B$ as I have done in my solution (i.e. they are events that correspond to the choice of coin), then the statement $\Pr[B \mid A]$ means "the probability that coin $B$ was selected, given that coin $A$ was selected." Obviously, that is $0$, and the formula says as much.