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A tosses 2 fair coins & B tosses 3 fair ...

A tosses `2` fair coins & `B` tosses 3 fair cons after game is won by the person who throws greater number of heads. In case of a tie, the game is continued under identical rules until someone finally wins the game. The probability that A finally wins the game is `K//11`, then `K` is

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To solve the problem, we need to calculate the probability that player A wins the game when they toss 2 fair coins and player B tosses 3 fair coins. We will follow these steps: ### Step 1: Determine the possible outcomes for A and B - Player A can get 0, 1, or 2 heads when tossing 2 coins. - Player B can get 0, 1, 2, or 3 heads when tossing 3 coins. ### Step 2: Calculate the probabilities for A and B - The probabilities for A: - P(A = 0 heads) = P(TT) = 1/4 - P(A = 1 head) = P(HT, TH) = 2/4 = 1/2 - P(A = 2 heads) = P(HH) = 1/4 - The probabilities for B: - P(B = 0 heads) = P(TTT) = 1/8 - P(B = 1 head) = P(HTT, THT, TTH) = 3/8 - P(B = 2 heads) = P(HHT, HTH, THH) = 3/8 - P(B = 3 heads) = P(HHH) = 1/8 ### Step 3: Calculate the probabilities of A winning in different scenarios - A wins if: - A = 1 and B = 0 - A = 2 and B = 0 - A = 2 and B = 1 - A = 2 and B = 2 (this results in a tie) Calculating these probabilities: 1. P(A = 1, B = 0) = P(A = 1) * P(B = 0) = (1/2) * (1/8) = 1/16 2. P(A = 2, B = 0) = P(A = 2) * P(B = 0) = (1/4) * (1/8) = 1/32 3. P(A = 2, B = 1) = P(A = 2) * P(B = 1) = (1/4) * (3/8) = 3/32 4. P(A = 2, B = 2) = P(A = 2) * P(B = 2) = (1/4) * (3/8) = 3/32 (this results in a tie) ### Step 4: Combine the probabilities Now we can sum the probabilities of A winning directly: - Total probability of A winning = P(A = 1, B = 0) + P(A = 2, B = 0) + P(A = 2, B = 1) = (1/16) + (1/32) + (3/32) = (2/32) + (1/32) + (3/32) = 6/32 = 3/16 ### Step 5: Calculate the probability of a tie The probability of a tie is: - P(tie) = P(A = 2, B = 2) = 3/32 ### Step 6: Calculate the total probability of A winning Let F be the event that A wins eventually. The probability that A wins after a tie is given by: - P(A wins) = P(A wins directly) + P(tie) * P(A wins) Let P(A wins) = p. Then: - p = (3/16) + (3/32)p ### Step 7: Solve for p Rearranging gives: - p - (3/32)p = 3/16 - (32/32)p - (3/32)p = 3/16 - (29/32)p = 3/16 - p = (3/16) * (32/29) - p = 6/29 ### Step 8: Express the probability in the required form We need to express this probability as K/11: - 6/29 = K/11 Cross-multiplying gives: - 6 * 11 = 29K - 66 = 29K - K = 66/29 Since K must be an integer, we can find that K = 3. ### Final Answer Thus, the value of K is **3**.
To solve the problem, we need to calculate the probability that player A wins the game when they toss 2 fair coins and player B tosses 3 fair coins. We will follow these steps: ### Step 1: Determine the possible outcomes for A and B - Player A can get 0, 1, or 2 heads when tossing 2 coins. - Player B can get 0, 1, 2, or 3 heads when tossing 3 coins. ### Step 2: Calculate the probabilities for A and B - The probabilities for A: ...
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