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Nutka1998 [239]

3 years ago

9

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game

, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

1 answer:

andrezito [222]3 years ago

7 0

Answer:

Probability is $\frac{1}{4}$

Step-by-step explanation:

This problem can be easily done tree diagram.

Few things keep in mind :

From starting of match three victories of A.
Less then or equal to two victories of B.
In each matches there are two possibilities either favourable for A or B.

First case if A wins first match

two victories of B then two of A
victory of B then A then B then A
victory of B then A then A

Second case if B wins first match(favorable)

victory of two B then three A

So total cases are 4 and favorable is one

probability= $\frac{1}{4}$

[Tree diagram is in attachment]

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Answer:

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Step-by-step explanation:

Step 1

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In this problem we have that

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<u>Find the area of the figure ABC</u>

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substitute

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$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

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The perimeter of the figure is equal to

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Applying Pythagoras theorem

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Remember that

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$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

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Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

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