Answer-first summary for fast verification
Answer: 2/5
## Explanation This is a conditional probability problem using Bayes' Theorem. We need to find P(Tails | Move 5 spaces). **Step 1: Define events** - Let T = event of flipping tails - Let H = event of flipping heads - Let M5 = event of moving 5 spaces **Step 2: Calculate P(M5 | T)** When tails is flipped, player rolls two dice. The total number of outcomes when rolling two dice is 6 × 6 = 36. The ways to get a sum of 5 with two dice are: (1,4), (2,3), (3,2), (4,1) → 4 favorable outcomes So P(M5 | T) = 4/36 = 1/9 **Step 3: Calculate P(M5 | H)** When heads is flipped, player rolls one die. The probability of rolling a 5 on a single die is: P(M5 | H) = 1/6 **Step 4: Calculate P(M5)** using law of total probability P(M5) = P(M5 | T) × P(T) + P(M5 | H) × P(H) P(M5) = (1/9) × (1/2) + (1/6) × (1/2) P(M5) = 1/18 + 1/12 = 2/36 + 3/36 = 5/36 **Step 5: Apply Bayes' Theorem** P(T | M5) = [P(M5 | T) × P(T)] / P(M5) P(T | M5) = [(1/9) × (1/2)] / (5/36) P(T | M5) = (1/18) / (5/36) = (1/18) × (36/5) = 36/(18×5) = 2/5 **Final Answer:** 2/5 **Verification:** The calculation matches the solution provided in the text: P(T/5) = (1/9) × (1/2) / (5/36) = 2/5
Author: Nikitesh Somanthe
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In a game, a coin is flipped. If the coin is heads, the player rolls one die. If the coin turns up tails, the player rolls two dice and the player moves their playing piece that number of spots shown on the die or dice. Given that on a player's turn, he moves 5 spaces, what is the probability he flipped tails on the coin?
A
1/10
B
1/5
C
2/5
D
1/3
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