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Hey there! I’m having trouble with questing 25c, any help would be appreciated.

Thankyou!

 Jul 23, 2018

Best Answer 

 #1
avatar+22884 
+1

Probability worded problem

Jane has three bags of lollies. In bag 1 there are three mints and three toffees,
in bag 2 there are three mints and two toffees and in bag 3 there are two mints and one toffee.
Jane selects a bag at random and then selects a lolly at random.

 

c the probability hat jane chose bag 1, given that she selects a mint.


\(\text{P(bag 1)$ = \dfrac13 $} \\ \text{P(mint/bag 1)$ = \dfrac36$} \\ \text{P(mint form bag 1)$ =$ P(bag 1)P(mint/bag 1)$ = \left(\dfrac13\right)\left(\dfrac36\right) = \dfrac16 $}\)

 

laugh

 Jul 23, 2018
 #1
avatar+22884 
+1
Best Answer

Probability worded problem

Jane has three bags of lollies. In bag 1 there are three mints and three toffees,
in bag 2 there are three mints and two toffees and in bag 3 there are two mints and one toffee.
Jane selects a bag at random and then selects a lolly at random.

 

c the probability hat jane chose bag 1, given that she selects a mint.


\(\text{P(bag 1)$ = \dfrac13 $} \\ \text{P(mint/bag 1)$ = \dfrac36$} \\ \text{P(mint form bag 1)$ =$ P(bag 1)P(mint/bag 1)$ = \left(\dfrac13\right)\left(\dfrac36\right) = \dfrac16 $}\)

 

laugh

heureka Jul 23, 2018
 #2
avatar+1698 
+1

Solution:

 

\(\text {For these types of question, use the Total Probability Theorem/Formula.}\\ \mathbb{P}(B_1) =\mathbb{P}(B_2) =\mathbb{P}(B_3) = \dfrac{1}{3}\\ \text {Probability of mint from Bag} \tiny \text {#} \\ \mathbb{P}(M|B_1) = \dfrac{1}{2}\\ \mathbb{P}(M|B_2) = \dfrac{3}{5}\\ \mathbb{P}(M|B_3) = \dfrac{2}{3}\\ \;\\ \mathbb{P}(B_1|M) = \Big (\mathbb{P}(B_1) \cdot \mathbb{P}(M|B_1)\Big)\div \Big(\mathbb{P}(B_1) \mathbb{P}(M|B1) + \mathbb{P}(B_2)\mathbb{P}(M|B_2) +\mathbb{P} P(B_3) \mathbb{P}(M|B_3)\Big)\)

 

\(\text {The probability that Jane chose bag 1, given that she selects a mint is equal to the probability}\\ \text {Jane chose a mint from bag #1 divided by the probability that she selected a mint.}\\\)

 

\(\mathbb{P}(B_1|M) = \dfrac{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) }{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) + \Big(\dfrac{1}{3} \cdot \dfrac{3}{5}\Big) +\Big( \dfrac{1}{3} \cdot \dfrac{2}{3}\Big)}\\ \;\\ \hspace{1.8cm}=\dfrac {15}{53}\\ \;\\ \hspace{1.8cm} \approx 0.28301\\ \)

 

 

GA

 Jul 24, 2018
edited by GingerAle  Jul 24, 2018

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