P(picking one defective) = 3/10
P(picking a 2nd defective) = 2/9
P(1 and 2 defective) = 3/10 x 2/9 = 6/90 = 0.066
Second method using combination:
³C₂ / ¹⁰C₂ = 1/15 = 0.066
The valid probability distributions are the ones in options C and D.
<h3>
Which of the following are valid probability distributions?</h3>
For discrete random variables with probabilities p₁, p₂, ..., pₙ, there are two rules:
- All of these probabilities are numbers between 0 and 1.
- p₁ + p₂ + ... + pₙ = 1.
So, for the first rule we can discard the first option, where we have negative probabilities.
To check the other 4 options, just add the probabilities and see if the addition gives 1.
The options that add up to 1 are C and D, so these two are the correct options.
D: 1/5 + 1/10 + 1/10 + 1/10 + 1/5 + 1/10 + 1/10 + 1/10 = 1
C: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1
If you want to learn more about probability:
brainly.com/question/25870256
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Answer: 3/2
Step-by-step explanation:
3/10 * 5
3/10 * 5/1
this can be simplified, the 5 cancels out and the 10 becomes a 2
3/2*1/1 = 3/2
First we rewrite the functions:
y = 2x
y = x ^ 10
We note that the second function always has values of y greater than the first function. However, there is a value of x for which the first function is greater.
For x = 1 we have:
y = 2 (1) = 2
y = (1) ^ 10 = 1
We note that:
2> 1
Answer:
Yes, the value of function y = 2x eventually exceed the value of function y = x ^ 10.
Answer:
30 students
Step-by-step explanation:
all students = girls+boys
all students= 15+35=50
40% of 50= 20
50-20=30
30 students didn't take the bus to school
