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:
brainly.com/question/24756209
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We need more information. what graph?
Answer:
jo pahle wale X hai na na X + x plus 1 unhen Yad Karke answer 1 aaega FIR X + 2 =_= + 3 X + 3 yah answer Aaya 1 or 2 X + ka ok
Step-by-step explanation:
FIR hai x+ x and ++ 2021 x
2021x
FIR + Kannan donon ko x
ans
x - 3 X
x - 4042
yah answer aayega ine donon ka the solve this question ok friend
