Troy worked for 3 hours and 15 minutes. How should he write that on his time card?
2 answers:
Answer:
Option B 3.25
Step-by-step explanation:
Troy worked for 3 hours and 15 minutes.
Since the timing on the card should be written on the card in terms of hours.
For which we will convert the time 3 hours 15 minutes into hours.
3 hours 15 minutes = 3 hours + 15 minutes
= 3 hours + hours
= 3.25 hours.
Option B 3.25 hours will be the answer.
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Without replacement from a small (8) number of balls means the binomial distribution is NOT applicable since the probability of each pick is not constant.
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
where 
= combination of r items selected from n,A+B=total number of items = 3+5 = 8a+b=number of items selected=4
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2