Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.
We are to find the probability of selecting 1 red apple and 2 yellow apples.
Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.
Then, we have

Therefore, the probability of event A is given by

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
First to Answer
Answer:
The correct option is the third option. -2 1/4 / -2/3 = 3 3/8
-9/4 / -2/3 = 3 3/8
Answer:
3
Step-by-step explanation:
I think it three sometimes it's a trick question
so its not number one or 2 because how are they her children so the answer is 3
A line segment has (three one two zero) endpoints
The answer is 2
You need to implement the process of "flip-flop" and multiply which is where you flip the second fraction and multiply.

÷

⇒

×

Now you multiply across. 2×15=30 and 3×11=33
So now it is

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To reduce to the lowest terms, find the GCF common factor
The GCF is 3
Divide both the numerator and denominator by 3

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=

Answer: