The probability of drawing two blue marbles if the first marble is not replaced is 1/5
<h3>How to determine the probabilities?</h3>
<u>The probability of tossing a head and drawing a red marble</u>
The given parameters are:
White = 1
Blue =3
Red = 2
Total = 6
The probability of a head is
P(Head)= 1/2
The probability of drawing a red marble is
P(Red)= 2/6 = 1/3
The required probability is
P = P(Head) * P(Red)
This gives
P = 1/2 * 1/3
P =1/6
<u>The probability of drawing two blue marbles if the first marble is not replaced.</u>
Here, we have:
P(B1) = 3/6 = 1/2
P(B2) = 2/5
The required probability is
P = P(B1) * P(B2)
This gives
P = 1/2 * 2/5
P =1/5
Hence, the probability of drawing two blue marbles if the first marble is not replaced is 1/5
Read more about probability at:
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He recognizes that the answer is whatever you can multiply by the second number in the division problem
Step-by-step explanation:
I think AL 76 my bad if it wrong
Answer:
5 cubed 5x5x5 5x5=25 25x5=125 answer 125
Step-by-step explanation:
