3. A bag contains five different lightbulbs. The bulbs are the same size but each has a different wattage. If you select three bulbs at a random,how many different outcomes will be in the sample space if the bulbs are selected without replacement?
I understand there is 5 different light bulbs, I tried to doing a tree diagraphm, but my answer was wrong and I am not sure if I am going in the right direction, the books answer is 4000. Could you explain to me how they got that answer.
i typed the wrong answer it is 125
i figured it out 5*5*5=125
To find the number of different outcomes in the sample space, we need to consider the number of possible ways to select three bulbs out of the five available bulbs without replacement.
The number of ways to choose three bulbs out of five without replacement can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the number of items to choose from, and r is the number of items to choose.
In this case, n = 5 (number of light bulbs) and r = 3 (number of bulbs to be selected). Plugging these values into the formula, we get:
C(5, 3) = 5! / (3!(5-3)!)
= 5! / (3!2!)
= (5*4*3*2*1) / ((3*2*1)*(2*1))
= (5*4) / (2*1)
= 10
Therefore, there are 10 different outcomes in the sample space when three light bulbs are selected without replacement.
The answer given in the book, 4000, seems to be incorrect.
To determine the number of different outcomes in the sample space, we need to consider the number of ways we can select 3 bulbs out of the 5 bulbs, without replacement.
One way to solve this problem is by using combinations. The number of combinations, denoted as nCr, gives us the number of ways we can choose r items from a set of n items, without regard to the order.
The formula for combinations is given by:
nCr = n! / (r! * (n-r)!)
In this case, we have 5 bulbs (n = 5) and want to select 3 bulbs (r = 3), so we can calculate the number of different outcomes using the formula:
5C3 = 5! / (3! * (5-3)!)
Simplifying this expression:
5! = 5 * 4 * 3 * 2 * 1 = 120
3! = 3 * 2 * 1 = 6
(5-3)! = 2!
Substituting these values back into our formula:
5C3 = 120 / (6 * 2)
5C3 = 120 / 12
5C3 = 10
So, there are 10 different outcomes in the sample space if you randomly select 3 bulbs without replacement.
It seems there's a discrepancy between the answer you provided (4000) and the correct answer (10). Please double-check the question or the book's solution, as there may have been a misinterpretation or a mistake.