Tree Diagram problems - what am i doing wrong?

During the Halloween Haunt Week of Kings Dominion, there is a carnival game where for $3 a ball, one can win a large teddy bear by shooting one ball into a straw basket several feet away.

A contestant has the choice to buy 3 balls (the maximum), 2 balls, 1 ball, or (of course) 0 balls (meaning that they are just watching other people play). After interviewing the carnival supervisor, it was found that 9% of their contestants buy three balls, 17% buy two balls, and 71% buy one ball. Of those that buy three balls, 70% win a teddy bear. Of those that buy two balls, 42% win a teddy bear. Of those that buy one ball, 15% win a teddy bear.

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Let's assume that these statistics are true. For the next person that arrives to the carnival booth, find the probability that he or she buys exactly one ball given that he or she wins a teddy bear.

i thought this one was just .71 x .15 = .1065 how is that wrong?


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For the next person that arrives to the carnival booth, find the probability that he or she did not buy three balls given that he or she did not win a teddy bear.

i thought this was (.17 x .58) + (.71 x .85) = .7021 how is that wrong?


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For the next person that arrives to the carnival booth, find the probability that he or she buys three balls or wins a teddy bear.

i didnt know how to do this one


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For the next person that arrives to the carnival booth, find the probability that he or she did not buy three balls given that he or she did not win a teddy bear.

i thought this was .7021 / .7291 but i got this wrong

what is the right answer?



thanks

Hello, bugmento!

During the Halloween Haunt Week of Kings Dominion, there is a carnival game where for $3 a ball,
one can win a large teddy bear by shooting one ball into a straw basket several feet away.

A contestant has the choice to buy 3 balls, 2 balls, or 1 ball.

It was found that 9% of their contestants buy three balls, 17% buy two balls, and 71% buy one ball.

Of those that buy three balls, 70% win a teddy bear.
Of those that buy two balls, 42% win a teddy bear.
Of those that buy one ball, 15% win a teddy bear.

Click to expand...

\(\displaystyle \text{We need Bayes' Theorem: }\;P(A\,|\,B) \;=\;\frac{P(A \wedge B)}{P(B)}\)

We can crank out some preliminary probabilities right now.

\(\displaystyle \begin{array}{ccccc}P(\text{3 balls }\wedge\text{ win}) &=& 0.09 \times 0.70 &=& 0.0630 \\ P(\text{2 balls }\wedge\text{ win}) &=& 0.17 \times 0.42 & = & 0.0714 \\ P(\text{1 ball }\wedge\text{ win}) &=& 0.71 \times 0.15 &=& 0.1065 \end{array}\)

\(\displaystyle \text{Hence: }\:\begin{array}{ccccc}P(\text{win}) &=& 0.063 + 0.0714 + 0.1065 &=& 0.2409 \\ P(\sim\text{win}) &=& 1 - 0.2409 &=& 0.7591 \end{array}\)

A man plays the game.

(1) Find the probability that he buys exactly one ball, given that he wins a teddy bear.

Click to expand...

\(\displaystyle \text{We want: }\;P(\text{1 ball }|\text{ win}) \;=\;\frac{P(\text{1 ball} \wedge\text{win})}{P(\text{win})} \;=\;\frac{0.1065}{0.2409} \;\approx\; 0.4421\)

(2) Find the probability that he did not buy three balls, given that he did not win a teddy bear.

Click to expand...

\(\displaystyle P(\sim\text{3 balls } \wedge \sim\text{win}) \;=\;\begin{Bmatrix}P(\text{1 ball }\wedge \sim\text{win}) &=& 0.71 \times 0.85 &=&0.6035 \\ P(\text{2 balls }\wedge \sim\text{win}) &=& 0.17 \times 0.58 &=& 0.0986 \end{Bmatrix} \;=\;0.7021\)

\(\displaystyle P(\sim\text{3 balls }|\sim\text{win}) \;=\;\frac{P(\sim\text{3 balls } \wedge \sim\text{win})}{P(\sim\text{win})} \;=\;\frac{0.7021}{0.7591} \;\approx\;0.9249\)

(3) Find the probability that he buys three balls or wins a teddy bear.

Click to expand...

\(\displaystyle \text{Formula: }\(A \vee B) \;=\;P(A) + P(B) - P(A \wedge B)\)

\(\displaystyle P(\text{3 balls }\vee\text{ win}) \;=\;P(\text{3 balls}) + P(\text{win}) - P(\text{3 balls }\wedge\text{ win})\)

. . . . . . . . . . . . .\(\displaystyle = \;0.09 + 0.2409 - 0.063 \;=\;0.2679\)

(4) Find the probability that he did not buy three balls, given that he did not win a teddy bear.

Click to expand...

This is identical to question (2) . . . a typo?

duplicate posT:

http://answers.yahoo.com/question/index ... 058AAhgKlh