3 bolts and 3 nuts are in a box. 2 parts are chosen at random. Find the probability that 1 is a bolt and 1 is a nut.
Why did the bolt and nut go to couples therapy? Because they couldn't screw things together anymore!
Now, to find the probability of getting 1 bolt and 1 nut out of 2 parts, let's calculate the total number of ways we can choose 2 parts from the box.
We have 6 parts in total (3 bolts + 3 nuts), so the total number of ways to choose 2 parts is given by the combination formula:
C(6, 2) = 6! / (2! * (6-2)!) = 15
Now, let's calculate the number of ways we can choose 1 bolt and 1 nut from the box.
We have 3 bolts and 3 nuts, so the number of ways to choose 1 bolt and 1 nut is the product of choosing 1 part from each category:
C(3, 1) * C(3, 1) = (3! / (1! * (3-1)!)) * (3! / (1! * (3-1)!)) = (3 * 3) = 9
Therefore, the probability of choosing 1 bolt and 1 nut out of 2 parts is:
9 / 15 = 3 / 5
So, the probability is 3/5, which means there's a pretty good chance you'll end up with a mixed pair!
To find the probability that one part is a bolt and one part is a nut, we need to determine the total number of ways we can choose 2 parts from the box.
The total number of ways to choose 2 parts from a box of 3 bolts and 3 nuts is given by the combination formula:
C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
So, there are 15 different ways we can choose 2 parts from the box.
Now, we need to determine the number of ways we can choose 1 bolt and 1 nut.
We can choose 1 bolt from the 3 available bolts in C(3, 1) = 3 ways, and choose 1 nut from the 3 available nuts in C(3, 1) = 3 ways.
Therefore, the number of ways to choose 1 bolt and 1 nut is 3 * 3 = 9.
Finally, the probability that 1 part is a bolt and 1 part is a nut is the number of favorable outcomes (9) divided by the total number of outcomes (15):
P = 9/15 = 3/5 = 0.6
So, the probability that 1 part is a bolt and 1 part is a nut is 0.6 or 60%.
To find the probability that 1 part is a bolt and 1 part is a nut, we need to determine the total number of possible outcomes and the number of successful outcomes.
Total Number of Outcomes:
There are 2 parts chosen at random from a box containing 3 bolts and 3 nuts. Since there are 6 parts in total, we can calculate the total number of outcomes by using the formula for combinations. The number of combinations of 6 parts taken 2 at a time is denoted by "C(6, 2)" and can be calculated as:
C(6, 2) = 6! / [(6-2)! * 2!] = (6 * 5 * 4 * 3!) / [(4 * 3!) * 2!] = (6 * 5) / 2 = 15
Therefore, the total number of possible outcomes is 15.
Number of Successful Outcomes:
In order to have 1 part selected as a bolt and 1 part selected as a nut, we need to choose 1 bolt out of 3 and 1 nut out of 3. The number of ways we can do this is the product of the number of ways to choose 1 bolt from 3 and the number of ways to choose 1 nut from 3. This can be calculated using the formula for combinations:
C(3, 1) * C(3, 1) = (3! / [(3-1)! * 1!]) * (3! / [(3-1)! * 1!]) = (3 * 2 * 1 * 3 * 2 * 1) / (1 * 2 * 1 * 1 * 2 * 1) = 6
Therefore, the number of successful outcomes is 6.
Calculating Probability:
The probability of an event is given by the ratio of the number of successful outcomes to the total number of possible outcomes.
Probability = Number of Successful Outcomes / Total Number of Outcomes = 6 / 15 = 2 / 5 = 0.4
Therefore, the probability that 1 part is a bolt and 1 part is a nut is 0.4 or 40%.
number of ways to choose 2 from 5 items = C(5,2) = 10
number of ways to choose 1 bolt from3 and 1 nut from 3
= C(3,1) * C(3,1) = 9
prob(your event) = 9/10