AMC 8 2000 Problem 21
AMC8 2000 年第 21 题
概率★★★☆☆
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
Keiko 掷了一枚便士，Ephraim 掷了两枚便士。Ephraim 得到的正面的个数和 Keiko 得到的正面的个数一样多的概率是：
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
Divide it into 2 cases: 1) Keiko and Ephriam both get 0 heads: This means that they both roll all tails, so there is only 1 way for this to happen. 2) Keiko and Ephriam both get 1 head: For Keiko, there is only 1 way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are 2 choices for when he can flip the head. So, in total there are ways for this case. Thus, in total there are 3 ways that work. Since there are 2 choices for each coin flip (Heads or Tails), there are total ways of flipping 3 coins. Thus, since all possible coin flips of 3 coins are equally likely, the probability is .
Let be the probability that Keiko gets heads, and let be the probability that Ephriam gets heads. , , K(2) = 0 (Keiko only has one penny!) , (because Ephraim can get HT or TH) The probability that Keiko gets 0 heads and Ephriam gets 0 heads is . Similarly for 1 head and 2 heads. Thus, we have: , , Thus the answer is .
Keiko can only get 1 head or one tail. That means Ephriam has to also get 1 head or one tail. There are 3 cases $HHT, HTH, TTT$ and each have a chance. Adding all 3 ways makes a total of .
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