To find the probability that the cube was rolled 6 times given that the coin landed on heads, we can use Bayes' theorem.
Let:
- \(A\) be the event that the cube was rolled 6 times.
- \(B\) be the event that the coin landed on heads.
We want to find \(P(A|B)\), which can be calculated using the formula:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
Step 1: Finding \(P(B|A)\)
This is the probability that the coin lands on heads given that the cube was rolled 6 times. From the given data, when the cube was rolled 6 times, there were 6 heads out of a total of (6 heads + 10 tails) = 16 total flips. So:
\[ P(B|A) = \frac{6}{16} = \frac{3}{8} \]
Step 2: Finding \(P(A)\)
This is the probability that the cube was rolled 6 times. From the total number of rolls (100), the number of times the cube was rolled 6 times is 16. So:
\[ P(A) = \frac{16}{100} = \frac{4}{25} \]
Step 3: Finding \(P(B)\)
This is the total probability of getting heads. We need to account for all the ways of rolling the cube and the respective heads observed.
Using the total counts:
- Heads when 1 roll: 9
- Heads when 2 rolls: 15
- Heads when 3 rolls: 12
- Heads when 4 rolls: 8
- Heads when 5 rolls: 9
- Heads when 6 rolls: 6
Adding these up gives the total number of heads over all rolls:
\[ \text{Total Heads} = 9 + 15 + 12 + 8 + 9 + 6 = 59 \]
Thus, the probability of getting heads is:
\[ P(B) = \frac{59}{100} \]
Step 4: Putting it all together
Now we can calculate \(P(A|B)\):
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{\left(\frac{3}{8}\right) \cdot \left(\frac{4}{25}\right)}{\frac{59}{100}} \]
Calculating the numerator:
\[ P(B|A) \cdot P(A) = \frac{3}{8} \cdot \frac{4}{25} = \frac{12}{200} = \frac{3}{50} \]
Now for \(P(A|B)\):
\[ P(A|B) = \frac{\frac{3}{50}}{\frac{59}{100}} = \frac{3}{50} \cdot \frac{100}{59} = \frac{6}{59} \]
Thus, the probability that the cube was rolled 6 times given that the coin landed on heads is:
\[ \boxed{\frac{6}{59}} \]