Respuesta :

Apple be A and blueberries be B

So

As per Venn diagram

  • n(A)=12
  • n(B)=7
  • n(A$\cap$B)=3

#1

  • n(A)+n(B)
  • 12+7
  • 19

P(A)=12/24

P(B)=7/24

Add

  • 12+7/24
  • 19/24

#2.

n(AUB)

  • n(A)+n(B)-n(An B).
  • 19-3
  • 16

E=8+16=24

P(AUB)=16/24=2/3

semsee45

Answer:

[tex]\sf a) \quad P(contains\:apple)+P(contains\:blueberry)=\dfrac{5}{6}[/tex]

[tex]\sf b) \quad P(contains\:apple\:or\:blueberry)=\dfrac{11}{15}[/tex]

c)    Not mutually exclusive events

Step-by-step explanation:

From inspection of the Venn diagram:

  • Total number of smoothies = 12 + 3 + 7 + 8 = 30
  • Number of smoothies containing Apple = 12 + 3 = 15
  • Number of smoothies containing Blueberry  = 3 + 7 = 10
  • Number of smoothies containing both Apple and Blueberry = 3
  • Number of smoothies not containing Apple or Blueberry = 8

[tex]\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}[/tex]

Let A = contains apple

Let B = contains blueberry

[tex]\implies\sf P(A)=\dfrac{15}{30}=\dfrac{1}{2}[/tex]

[tex]\implies\sf P(B)=\dfrac{10}{30}=\dfrac{1}{3}[/tex]

Part (a)

[tex]\begin{aligned} \implies \sf P(A)+P(B) & =\sf \dfrac{1}{2}+\dfrac{1}{3}\\\\ & = \sf \dfrac{3}{6}+\dfrac{2}{6}\\\\ & = \sf \dfrac{5}{6}\end{aligned}[/tex]

Part (b)

[tex]\textsf{Addition Law}: \quad\sf P(A\:or\: B) = P(A)+P(B)-P(A\:and\:B)[/tex]

Given:

[tex]\sf P(A)+P(B)=\dfrac{5}{6} \quad \textsf{(from part a)}[/tex]

[tex]\sf P(A\:and\:B)=\dfrac{3}{30}\quad \textsf{(where the circles overlap)}[/tex]

[tex]\begin{aligned}\implies \sf \sf P(A\:or\: B) &= \sf P(A)+P(B)-P(A\:and\:B)\\\\& =\sf \dfrac{5}{6}-\dfrac{3}{30}\\\\ & =\sf \dfrac{25}{30}-\dfrac{3}{30}\\\\ & =\sf \dfrac{22}{30}\\\\ & = \sf \dfrac{11}{15}\end{aligned}[/tex]

Or, we can simply read P(contains apple or blueberry) from the Venn diagram.  

P(A or B) is the total of the numbers inside the circles divided by the total number of smoothies:

[tex]\sf P(A\:or\:B) = \dfrac{12+3+7}{30}=\dfrac{22}{30}=\dfrac{11}{15}[/tex]

Part (c)

For two events, A and B, where A and B are mutually exclusive:

[tex]\sf P(A \: or \: B)=P(A)+P(B)[/tex]

Given:

[tex]\sf P(A)+P(B)=\dfrac{5}{6} \quad \textsf{(from part a)}[/tex]

[tex]\sf P(A\:or\:B)=\dfrac{11}{15} \quad \textsf{(from part b)}[/tex]

[tex]\sf As\:\dfrac{11}{15}\neq \dfrac{5}{6} \implies P(A \: or \: B)\neq P(A)+P(B)[/tex]

Therefore, choosing a smoothie containing apple and choosing a smoothie containing blueberry are NOT mutually exclusive events.

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