You have a bag of enough ping pong balls such that each one has exactly one of the numbers from 57 to 133 painted on it. What is the probability that if you randomly select five of the balls from the bag, all of the numbers on the selected balls will be divisible by 3 or by 7? Your final answer must be a decimal number rounded to three decimal places. For example, if you calculate the answer 0.166666, your answer would be 0.167.
The Correct Answer and Explanation is:
To solve this problem, we need to determine the probability that five randomly selected ping pong balls have numbers that are divisible by 3 or 7. The total range of numbers on the balls is from 57 to 133.
Step 1: Find the total number of balls
The total number of ping pong balls is determined by counting the numbers from 57 to 133. This can be calculated as:133−57+1=77133 – 57 + 1 = 77133−57+1=77
So, there are 77 ping pong balls in total.
Step 2: Count the numbers divisible by 3 or 7
Now, we need to find how many numbers in the range 57 to 133 are divisible by 3, by 7, and by both.
Numbers divisible by 3
To find the numbers divisible by 3, we start by finding the first and last numbers in the range that are divisible by 3:
- The first number divisible by 3 is 57 (since 57 ÷ 3 = 19).
- The last number divisible by 3 is 132 (since 132 ÷ 3 = 44).
The numbers divisible by 3 are 57, 60, 63, …, 132. These numbers form an arithmetic sequence with the first term 57, the common difference 3, and the last term 132. The number of terms in this sequence is:132−573+1=26\frac{132 – 57}{3} + 1 = 263132−57+1=26
So, there are 26 numbers divisible by 3.
Numbers divisible by 7
Now, find the numbers divisible by 7 in the same way:
- The first number divisible by 7 is 56 (since 56 ÷ 7 = 8).
- The last number divisible by 7 is 133 (since 133 ÷ 7 = 19).
The numbers divisible by 7 are 56, 63, 70, …, 133. These also form an arithmetic sequence with the first term 56, the common difference 7, and the last term 133. The number of terms in this sequence is:133−567+1=12\frac{133 – 56}{7} + 1 = 127133−56+1=12
So, there are 12 numbers divisible by 7.
Numbers divisible by both 3 and 7
Next, find the numbers divisible by both 3 and 7, i.e., divisible by 21. The first number divisible by 21 is 63, and the last number divisible by 21 is 126. The numbers divisible by 21 are 63, 84, 105, …, 126. These numbers form an arithmetic sequence with the first term 63, the common difference 21, and the last term 126. The number of terms is:126−6321+1=4\frac{126 – 63}{21} + 1 = 421126−63+1=4
So, there are 4 numbers divisible by both 3 and 7.
Apply the Inclusion-Exclusion Principle
Using the inclusion-exclusion principle, the number of numbers divisible by 3 or 7 is:26 (divisible by 3)+12 (divisible by 7)−4 (divisible by both 3 and 7)=3426 \text{ (divisible by 3)} + 12 \text{ (divisible by 7)} – 4 \text{ (divisible by both 3 and 7)} = 3426 (divisible by 3)+12 (divisible by 7)−4 (divisible by both 3 and 7)=34
So, there are 34 numbers divisible by 3 or 7.
Step 3: Calculate the probability
The probability that a randomly selected ball has a number divisible by 3 or 7 is the ratio of favorable outcomes (34) to the total outcomes (77):Probability=3477≈0.4416\text{Probability} = \frac{34}{77} \approx 0.4416Probability=7734≈0.4416
Step 4: Calculate the probability for five balls
For five balls to all have numbers divisible by 3 or 7, we need to calculate the probability of selecting five balls, each with a favorable number. The probability that each ball selected is divisible by 3 or 7 is 0.44160.44160.4416, and since the selections are independent, the probability for five balls is:0.44165≈0.02950.4416^5 \approx 0.02950.44165≈0.0295
Final Answer:
The probability that five randomly selected balls all have numbers divisible by 3 or 7 is approximately 0.030 when rounded to three decimal places.
