We initially set
𝜃
𝐴
=
0.1
and
𝜃
𝐵
=
0.6
. What is the likelihood of obtaining 6 heads with coin A?

Question

We initially set 
𝜃
𝐴
=
0.1
 and 
𝜃
𝐵
=
0.6
 . What is the likelihood of obtaining 6 heads with coin A?

BlackTom AI · Accepted Answer

Let's restate the problem to ensure clarity: we have two coins with initial head-probabilities θ_A = 0.1 and θ_B = 0.6, and we are asked for the likelihood of obtaining 6 heads with coin A.

First, interpret the scenario for coin A. If θ_A represents the probability of heads on a single flip of coin A, then obtaining exactly 6 heads (presumably in 6 flips) follows a Bernoulli/binomial model. If we flip coin A 6 times and want the probability that all 6 results are heads, the probability is (θ_A)^6 = (0.1)^6.

Compute (0.1)^6: 0.1 × 0.1 × 0.1 × 0.1 × 0.1 × 0.1 = 1e-6 = 0.000001.

Now compare with each option:
- Option 0.71: This is vastly larger than the computed 0.000001, so it is not a match. A misinterpretation here might treat the event as highly probable, but with θ_A = 0.1, six heads in a row is extremely unlikely.
- Option None of these options are correct.: Since 0.000001 is not equal to any of the given numeric options (0.71, 0.11, 0.58, 0.005), this option remains plausible as the correct choice if none of the listed values matches the true likelihood.
- Option 0.11: This is off by several orders of magnitude from 0.000001; a common mistake could be misreading as a rough binomial probability or mixing up a different scenario (e.g., six or more heads, or a different number of trials).
- Option 0.58: This value is far from the actual probability and could arise from misapplying a rule or mixing up θ_B with θ_A, or considering a large-sample approximation improperly.
- Option 0.005: While closer in scale to 0.000001 than some other options, 0.005 is still 500 times larger than the correct 1e-6, suggesting a miscalculation such as treating the event as six heads in a single trial count different from six flips, or misplacing decimal places.

In summary, the direct calculation for the probability of six consecutive heads with a coin that has head-probability θ_A = 0.1 yields (0.1)^6 = 1e-6, which matches none of the numeric options provided. Therefore, the only option that remains logically consistent with the calculation is the one that states that none of the listed options are correct.

25S-STATS-102B-LEC-3 Quiz 8- Requires Respondus LockDown Browser

We initially set 𝜃 𝐴 = 0.1 and 𝜃 𝐵 = 0.6 . What is the likelihood of obtaining 6 heads with coin A?

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