Question

An urn contains 5 green balls, 8 red balls and 7 blue balls. Event A is defined as drawing a red ball on the first draw, and event B is defined as drawing a red ball on the second draw. If two balls are drawn from the urn, one after the other and not replaced, what is P(B|A) expressed in simplest form?

Answer 1

Answer:

no.of green balls = 5

no.of red balls = 8

no.of blue balls = 7

total no.of balls = 5+8+7

= 20

Answer 2

$\displaystyle \bf P(B|A) = \frac{7}{19}$

Given :

• An urn contains 5 green balls, 8 red balls and 7 blue balls.

• Event A is defined as drawing a red ball on the first draw, and event B is defined as drawing a red ball on the second draw.

• Two balls are drawn from the urn, one after the other and not replaced

To find :

P(B|A) in simplest form

Solution :

Step 1 of 2 :

Find P(A) , P(B) and P(A ∩ B)

An urn contains 5 green balls, 8 red balls and 7 blue balls

Total number of balls = 5 + 8 + 7 = 20

Event A is defined as drawing a red ball on the first draw, and event B is defined as drawing a red ball on the second draw

Two balls are drawn from the urn, one after the other and not replaced

P(A)

= The probability of drawing a red ball on the first draw

$\displaystyle \sf = \frac{8}{20}$

$\displaystyle \sf = \frac{2}{5}$

P(B)

= The probability of drawing a red ball on the second draw

$\displaystyle \sf = \frac{7}{19}$

Now , A and B are two mutually independent events

$\displaystyle \sf \therefore \: P(A \cap B)$

$\displaystyle \sf = P(A) \times P(B)$

$\displaystyle \sf = \frac{1}{5} \times \frac{7}{19}$

$\displaystyle \sf = \frac{7}{95}$

Step 2 of 2 :

Find P(B|A) in simplest form

$\displaystyle \sf P(B|A)$

$\displaystyle \sf = \frac{ P(A \cap B)}{P(A)}$

$\displaystyle \sf = \frac{ \frac{7}{95} }{ \frac{1}{5} }$

$\displaystyle \sf = \frac{7}{95} \times \frac{5}{1}$

$\displaystyle \sf = \frac{7}{19}$

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