Answer:
Binomial distribution requires all of the following to be satisfied:
1. size of experiment (N=27) is known.
2. each trial of experiment is Bernoulli trial (i.e. either fail or pass)
3. probability (p=0.14) remains constant through trials.
4. trials are independent, and random.
Binomial distribution can be used as a close approximation, with the usual assumption that a sample of 27 in thousands of stock is representative of the population., and is given by the probability of x successes (defective).
P(x)=C(N,x)*p^x*(1-p)^(n-x)
where N=27, p=0.14, and C(N,x) is the number of combinations of x items out of N.
So we need the probability of <em>at most one defective</em>, which is
P(0)+P(1)
= C(27,0)*0.14^0*(0.86)^(27) + C(27,1)*0.14^1*(0.86^26)
=1*1*0.0170 + 27*0.14*0.0198
=0.0170+0.0749
=0.0919
Hi Jessica,
<span>Remember PEMDAS (Parenthesis, Exponents, Multiplication & Division, Addition & Subtraction).
√3 x 66.15/4.41 {Exponents/Cube & Square Roots First}
1.73 x 66.15 ÷ 4.41 {Multiplication}
114.4395 ÷ 4.41 {Division}
25.95 {Final Answer}
Cheers,
Izzy</span>
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I think that the answer is 120 cubes
Answer:
The probability of the event that first ball that is drawn is blue is
.
Step-by-step explanation:
Probability:
If S is is an sample space in which all outcomes are equally likely and E is an event in S, then the probability of E,denoted P(E) is

Given that,
An urn contains two balls B₁ and B₂ which are blue in color and W₁,W₂ and W₃ which are white in color.
Total number of ball =(2+3) =5
The number ways of selection 2 ball out of 5 ball is
=5²
=25
Total outcomes = 25
List of all outcomes in the event that the first ball that is drawn is blue are
B₁B₁ , B₁B₂ , B₁W₁ , B₁W₂ , B₁W₃ , B₂B₁ , B₂B₂ , B₂W₁ , B₂W₂ , B₂W₃
The number of event that the first ball that is drawn is blue is
=10.
The probability of the event that first ball that is drawn is blue is

