257. Some toys include large, middle, and small model with red, yellow, green, or blue color. If numbers of all model-color combinations are the same, for example, number of red large toys is equal to number of green little toys. A boy wants a red-large toy. If his mother select one for him at random, what is the probability that at least one of the color and model will satisfy the boy?
The question stem seems to be asking:varun289 wrote:257. Some toys include large, middle, and small model with red, yellow, green, or blue color. If numbers of all model-color combinations are the same, for example, number of red large toys is equal to number of green little toys. A boy wants a red-large toy. If his mother select one for him at random, what is the probability that at least one of the color and model will satisfy the boy?
If the boy wants a toy that is BOTH large AND red, what is the probability that he gets a toy with AT LEAST ONE of these two desired qualities?
Thus, a favorable outcome will be achieved if the selected toy is large, red, or BOTH large AND red.
This problem is not only about probability but also about OVERLAPPING GROUPS.
Toys that are large OR red = (large toys) + (red toys) - (toys that are BOTH large AND red).
The big idea with overlapping group problems is to SUBTRACT THE OVERLAP.
When we count the number of large toys and the number of red toys, the number of toys that are BOTH large AND red -- the OVERLAP -- gets counted twice.
So that we don't double-count the overlap, it must be SUBTRACTED from the total.
It is given that the toys include the same number of each size-color combination.
To make the math easier, let there be exactly one of each size-color combination, for a total of 12 toys.
Number of large toys = 4. (One of each color.)
Number of red toys = 3. (Small, medium and large.)
Number of toys that are BOTH large AND red = 1.
Thus:
Toys that are large OR red = 4 + 3 - 1 = 6.
Thus:
P(selecting a toy that is large OR red) = 6/12 = 1/2.
