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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 Head: 0.9**

** Probability of Tail: 0.1**

**Coin B –**

** Probability of Head: 0.1**

** Probability of Tail: 0.9**

************************************************************************************************

**Event A: Coin A is chosen**

**Event B: first 10 tosses are heads**

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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.

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.