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Alenkinab
2 months ago
6

P(vanilla) =0.3 P(sundae)=0.2 P(vanilla and sundae) =0.15 find the probability that a customer ordered vanilla ice cream given t

hey ordered a sundae. P(vanilla | sundae) =?
Mathematics
1 answer:
Svet_ta [12.7K]2 months ago
6 0

Answer:

0.75

Step-by-step explanation:

P(A | B) = P(A and B) / P(B)

P(vanilla | sundae) = P(vanilla and sundae) / P(sundae)

P(vanilla | sundae) = 0.15 / 0.2

Thus, P(vanilla | sundae) = 0.75

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James knows that when h walks, he takes about 120 steps per minute and that each step is 2.75 feet long. Which calculation is mo
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Answer:

120*2.75*60/5280

= 3.75

Step-by-step explanation:

Considering James takes 120 steps in a minute, with 60 minutes in an hour and a mile consisting of 5,280 feet, we can establish his walking pace. By multiplying his steps per minute by the distance of each step (120 * 2.75), we determine the distance he covers in one minute. This value is then multiplied by 60 to account for the total hour. Finally, the total distance is divided by the feet in a mile (5,280), which results in a speed of 3.75 miles per hour. Thus, the calculation becomes 120*2.75*60/5280

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Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.
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A="b is situated in the center"

B="c lies to the right of b"

C="The letters def occur sequentially in that arrangement"

a) b can occupy 7 positions; however, only one of these is the center. Therefore, P(A)=1/7

b) Let X=i; "b holds the i-th position"

Y=j; "c occupies the j-th position"

P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}

P(B)=1/2

c) Let X=i; "d holds the i-th position"

Y=j; "e occupies the j-th position"

Let Z=k; "f is in the i-th position"

P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420

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P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

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Events A and B are independent

P(A∩C)=1/420

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Events A and C are not independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

Events B and C are not independent

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