设 $X$ 是投掷距离，则 $X\sim N(40,4^2)$。标准化 $38.8$ 得到 $z=-0.3$，再用标准正态表。

Let $X\sim N(40,4^2)$.

Then

这是条件概率。已知已经 qualified，要求“第二投才 qualified 且第二投超过 $44$”。第二投才 qualified 表示第一投没有超过 $38.8$，第二投超过 $44$。分母是三次内至少一次超过 $38.8$。

From part (a),

So the probability of qualifying in at most $3$ attempts is

Also,

The probability that the first throw does not qualify and the second throw is more than $44$ m is

Therefore

So the probability is

直接概率相加法。对于分母 $P(\text{qualify})$（运动员在三次尝试中成功晋级决赛的概率），我们可以将其分解为三个互斥事件的和：

1. 第一投就成功晋级（$P(X_1 > 38.8)$）
2. 第一投失败，第二投成功晋级（$P(X_1 < 38.8 \cap X_2 > 38.8)$）
3. 前两投失败，第三投成功晋级（$P(X_1 < 38.8 \cap X_2 < 38.8 \cap X_3 > 38.8)$） 分别计算这三个概率后相加，可以得到与补集法完全相同的晋级概率。这种方法更为直观地描述了比赛的进程。

Let $p_F = P(X < 38.8) = 0.3821$ be the probability of failing to qualify on any single throw, and $p_Q = P(X > 38.8) = 1 - 0.3821 = 0.6179$ be the probability of qualifying on a single throw.

Calculate the probability of qualifying in at most $3$ attempts as the sum of three mutually exclusive paths:

• Qualify on 1st throw: $P(\text{1st}) = p_Q = 0.6179$
• Qualify on 2nd throw (1st fail, 2nd qualify): $P(\text{2nd}) = p_F \cdot p_Q = 0.3821 \times 0.6179 \approx 0.236099$
• Qualify on 3rd throw (1st and 2nd fail, 3rd qualify): $P(\text{3rd}) = p_F^2 \cdot p_Q = (0.3821)^2 \times 0.6179 \approx 0.090213$

Summing these paths:

Now calculate the probability of qualifying on the 2nd throw with a distance of more than $44$ m:

Finally, calculate the conditional probability:

So the probability is $0.0642$ (to $3$ significant figures).