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Sagot :
Answer:Let's clarify and break down the problem statement for a clearer understanding.
You have a bag containing 40 tickets numbered from 1 to 40. Four tickets are drawn at random and arranged in ascending order. The task is to find the probability that the number 25 is among the four tickets drawn.
To solve this, we follow these steps:
1. **Calculate the Total Number of Ways to Draw 4 Tickets Out of 40:**
The number of ways to draw 4 tickets from 40 without considering the order is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of tickets, and \( k \) is the number of tickets to be drawn:
\[
\binom{40}{4} = \frac{40!}{4!(40-4)!} = \frac{40!}{4! \cdot 36!}
\]
2. **Calculate the Number of Favorable Outcomes:**
If we want the number 25 to be included among the 4 tickets drawn, we need to select 3 more tickets out of the remaining 39 tickets (since 25 is already chosen).
The number of ways to draw 3 tickets from the remaining 39 is given by:
\[
\binom{39}{3} = \frac{39!}{3!(39-3)!} = \frac{39!}{3! \cdot 36!}
\]
3. **Calculate the Probability:**
The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes:
\[
P(\text{25 is drawn}) = \frac{\binom{39}{3}}{\binom{40}{4}}
\]
Now, let's compute these values:
\[
\binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390
\]
\[
\binom{39}{3} = \frac{39 \times 38 \times 37}{3 \times 2 \times 1} = 9,139
\]
Therefore, the probability is:
\[
P(\text{25 is drawn}) = \frac{9,139}{91,390} = \frac{9139}{91390} = \frac{1}{10}
\]
So, the probability that the number 25 is among the four tickets drawn is \( \frac{1}{10} \) or 0.1.
Step-by-step explanation:
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