Question 1200375: Raffle tickets numbered 1 through 30 are placed in a box. Tickets for a second raffle are
numbered 21 through 48 and placed in a second box. One ticket is randomly drawn from each box.
Find the probability the first ticket is greater than 10, and the second ticket is less than 40 or odd
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! Certainly, let's find the probability.
**1. Probability of the first ticket being greater than 10:**
* There are 20 tickets numbered greater than 10 (from 11 to 30) out of a total of 30 tickets.
* Probability of the first ticket being greater than 10: 20/30 = 2/3
**2. Probability of the second ticket being less than 40 or odd:**
* **Less than 40:** There are 19 tickets less than 40 (from 21 to 39) out of 28 total tickets.
* **Odd numbers:** There are 14 odd numbers (21, 23, 25, ..., 47) out of 28 total tickets.
* **Less than 40 and odd:** There are 10 numbers that are both less than 40 and odd (21, 23, 25, ..., 39).
* To find the probability of "less than 40 or odd", we use the formula:
* P(A or B) = P(A) + P(B) - P(A and B)
* where:
* P(A) = Probability of the ticket being less than 40
* P(B) = Probability of the ticket being odd
* P(A and B) = Probability of the ticket being both less than 40 and odd
* P(less than 40 or odd) = (19/28) + (14/28) - (10/28) = 23/28
**3. Probability of both events happening:**
* Since the draws are independent, the probability of both events happening is the product of their individual probabilities.
* Probability = (Probability of first ticket > 10) * (Probability of second ticket < 40 or odd)
* Probability = (2/3) * (23/28)
* Probability = 23/42
**Therefore, the probability that the first ticket is greater than 10 and the second ticket is less than 40 or odd is 23/42.**
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