you have 8 marbles numbered 1-8, mix them up & draw 2. What are the chances that the #s are 4 & 7
2/8 * 1/7
the first marble is four or seven=2*1/8
the second marble is the the other number: 1/7
Pr= 2*1/8*1/7=
To calculate the chances of drawing the numbers 4 and 7 from a set of 8 marbles, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's calculate the total number of possible outcomes. When drawing two marbles out of 8, we can arrange them in a total of 8C2 ways, which is calculated using the combination formula:
nCr = n! / [(n-r)! * r!]
In our case, n (the total number of marbles) is 8, and r (the number of marbles drawn) is 2. Plugging these values into the formula, we get:
8C2 = 8! / [(8-2)! * 2!]
= 8! / [6! * 2!]
= (8 * 7 * 6!) / (6! * 2)
= (8 * 7) / 2
= 56 / 2
= 28
Therefore, there are 28 possible outcomes when drawing two marbles from a set of 8.
Next, let's calculate the number of favorable outcomes, which is the number of ways we can draw the numbers 4 and 7. Since we are drawing without replacement, we can only draw each marble once.
To draw the number 4, we have one favorable marble out of 8, and then there are 7 remaining marbles. To draw the number 7, we have one favorable marble out of the remaining 7. Therefore, the number of favorable outcomes is:
1 favorable way to draw number 4 * 1 favorable way to draw number 7 = 1 * 1 = 1
So, there is only 1 favorable outcome - drawing both the number 4 and the number 7.
Finally, we can calculate the chances (probability) by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes
= 1 / 28
Therefore, the chances or probability of drawing the numbers 4 and 7 from a set of 8 marbles is 1/28.