Suppose you have 3 jars with the following contents. Jar 1 has 1 white ball and 4 black balls. Jar 2 has 2 white balls and 1 black ball. Jar 3 has 3 white balls and 2 black balls. One jar is to be ​selected, and then 1 ball is to be drawn from the selected jar. The probabilities of selecting the​ first, second, and third jars are​ 1/2​, ​1/3​, and ​1/6 respectively. Find the probability the ball was drawn from Jar 1​, given that the ball is white.

Respuesta :

Answer:

The probability is:

              [tex]\dfrac{9}{38}[/tex]

Step-by-step explanation:

We need to use the Baye's theorem in order to find the probability .

Jar 1:  has 1 white ball and 4 black balls.

This means that the probability of white ball is: 1/5

( since there are a total of 1+4=5 balls out of which 1 is white)

Jar 2: has 2 white balls and 1 black ball.

This means that the probability of white ball is: 2/3

( since there are a total of 2+1=3 balls out of which 2 are white)

Jar 3 : has 3 white balls and 2 black balls.

This means that the probability of white ball is: 3/5

( since there are a total of 3+2=5 balls out of which 3 are white)

Hence, the probability the ball was drawn from Jar 1​, given that the ball is white is:

Ratio of drawing jar 1 and a white ball from it to the sum of choosing each jar and a white ball from it.

i.e.

[tex]=\dfrac{\dfrac{1}{2}\times \dfrac{1}{5}}{\dfrac{1}{2}\times \dfrac{1}{5}+\dfrac{1}{3}\times \dfrac{2}{3}+\dfrac{1}{6}\times \dfrac{3}{5}}\\\\\\=\dfrac{\dfrac{1}{10}}{\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{1}{10}}\\\\\\=\dfrac{\dfrac{1}{10}}{\dfrac{2}{10}+\dfrac{2}{9}}\\\\\\=\dfrac{\dfrac{1}{10}}{\dfrac{38}{90}}\\\\\\=\dfrac{9}{38}[/tex]