Disk of ploycarbonate plastic form a supplier are anaylzed for scratcha nd shock resistance. the results from 100 disks are summarized below:
Shock Resistance
high low
Scratch high 70 9
resistance low 16 5
Let A denote the even that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Determine the number of disks in: P(A), P(B), P(A'),P(A and B), P(A union B), P(A' union B).
Ah, numbers, my arch-nemesis! But fear not, for Clown Bot is here to make statistics fun! Let's dive into this analysis with a dash of humor:
To determine the number of disks in each probability, we can use the given information. Brace yourself, here comes some math comedy!
First, let's calculate the number of disks in P(A), which represents the probability of having high shock resistance.
P(A) = (Number of disks with high shock resistance) / (Total number of disks)
= (70 + 9) / 100
= 79 / 100
So, the number of disks in P(A) is approximately 79. Now, let's move on to the next act!
Next up, we have P(B), which represents the probability of having high scratch resistance.
P(B) = (Number of disks with high scratch resistance) / (Total number of disks)
= (70 + 16) / 100
= 86 / 100
Hence, the number of disks in P(B) is around 86. Keep those probabilities coming!
Now, let's talk about P(A'), which represents the probability of not having high shock resistance. In other words, it's like the "opposite day" for high shock resistance.
P(A') = 1 - P(A)
= 1 - (79 / 100)
= 21 / 100
So, there are approximately 21 disks in P(A'). They may not be good at handling shocks, but they'll bounce back with a smile!
Now, let's team up A and B! P(A and B) refers to the probability of having both high shock resistance and high scratch resistance.
P(A and B) = (Number of disks with high shock resistance and high scratch resistance) / (Total number of disks)
= 70 / 100
= 7 / 10
Tada! We have around 7 disks in P(A and B).
Moving on to the next act, we have P(A union B), which represents the probability of having either high shock resistance or high scratch resistance or both. It's like a magical circus where everyone is welcome!
P(A union B) = P(A) + P(B) - P(A and B)
= (79 / 100) + (86 / 100) - (7 / 10)
= (79 + 86 - 70) / 100
= 95 / 100
Drumroll, please! P(A union B) has around 95 disks. Quite a crowd, wouldn't you say?
And finally, we have P(A' union B), which is the probability of not having high shock resistance or having high scratch resistance or both. It's a mix of twists and turns!
P(A' union B) = 1 - P(A union B)
= 1 - (95 / 100)
= 5 / 100
In P(A' union B), we find approximately 5 disks. They might have missed out on high shock resistance, but they've got high scratch resistance to make up for it!
Voila! We've analyzed the probabilities and identified the number of disks in each scenario. I hope I've added a humorous twist to this statistical circus!
To determine the probabilities in this scenario, we need to understand some basic concepts of probability theory.
1. P(A) represents the probability of event A occurring.
2. P(B) represents the probability of event B occurring.
3. P(A') represents the probability of the complement of event A (not-A or event A not occurring).
4. P(A and B) represents the probability of events A and B both occurring.
5. P(A union B) represents the probability of events A or B (or both) occurring.
6. P(A' union B) represents the probability of the complement of event A or event B occurring.
Let's calculate the probabilities step by step:
1. P(A): To determine the number of disks with high shock resistance, we add up the frequencies of high shock resistance in the table: 70 + 9 = 79. So, there are 79 disks with high shock resistance out of 100. Hence, P(A) = 79/100 = 0.79.
2. P(B): To determine the number of disks with high scratch resistance, we add up the frequencies of high scratch resistance in the table: 70 + 16 = 86. So, there are 86 disks with high scratch resistance out of 100. Hence, P(B) = 86/100 = 0.86.
3. P(A'): To find the complement of event A (not-A or event A not occurring), we subtract P(A) from 1: P(A') = 1 - P(A) = 1 - 0.79 = 0.21.
4. P(A and B): To determine the number of disks with both high shock resistance and high scratch resistance, we look at the intersection of the two categories in the table: 70. So, there are 70 disks with both high shock resistance and high scratch resistance out of 100. Hence, P(A and B) = 70/100 = 0.70.
5. P(A union B): To find the probability of either event A or event B (or both) occurring, we need to add the probabilities of each event individually and subtract the probability of their intersection to avoid double-counting: P(A union B) = P(A) + P(B) - P(A and B) = 0.79 + 0.86 - 0.70 = 0.95.
6. P(A' union B): To find the probability of the complement of event A or event B occurring, we add the probabilities of the complement of event A (P(A')) and event B (P(B)) and subtract the probability of their intersection to avoid double-counting: P(A' union B) = P(A') + P(B) - P(A and B) = 0.21 + 0.86 - 0.70 = 0.37.
Therefore, the number of disks in:
- P(A) = 0.79 (or 79 disks out of 100).
- P(B) = 0.86 (or 86 disks out of 100).
- P(A') = 0.21 (or 21 disks out of 100).
- P(A and B) = 0.70 (or 70 disks out of 100).
- P(A union B) = 0.95 (or 95 disks out of 100).
- P(A' union B) = 0.37 (or 37 disks out of 100).