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The above-given problem is from MIT-OCW probability – https://tinyurl.com/25wamnxz

There are two biased coins –  A and B. The probability of choosing either coin is 0.5. Once the coins are chosen, we perform the experiment of tossing the coins as shown in the figure above. The probabilities for heads and tails for respective coins are given below:

Coin A –

Probability of Tail: 0.1

Coin B –

Probability of Tail: 0.9

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Event A: Coin A is chosen

Event B: first 10 tosses are heads

### What is the probability of (A | B) i.e. if we get 10 successive heads, then what is the probability that coin A was chosen?

C be the event that coin B was chosen

P(A | B) = P(B | A) *(P(A) / P(B))

P(B) = P(B | A) * P(A) + P(B | C)*P(C)

= $(0.9)^{10} * 0.5 + (0.1)^{10} * 0.5$

= 0.17443392201 approx.

Hence

P(A | B) = $(0.9)^{10} * 0.5 / ((0.9)^{10} * 0.5 + (0.1)^{10} * 0.5)$

= 0.999457 Approx.
by

### 1 comment

edited
Sorry, I had posted the wrong question. I have modified the question. Please have a look.