#### Q. No. 5 :

Box u = 1 contains 1 red ball and 3 black balls. Box u = 2 contains 1 red ball, 1 white ball, and 1 black ball. Box u = 3 contains 1 red ball and 1 black ball.

(a) A box is chosen at random and one ball is drawn. What is the probability that the ball is red?

(a) A box is chosen at random and one ball is drawn. What is the probability that the ball is red?

(b) Given that the ball is red, what are the probabilities that the chosen box was box u = 1, 2, 3?

#### Answer :

#### (a) A box is chosen at random and one ball is drawn. What is the probability that the ball is red?

There are 3 boxes.

The probability of choosing any one box is 1/3.

There are 4 balls in box u=1.

Among these 4 balls 1 ball is red.

The probability of drawing a red ball from box u=1 is 1/4.

Among all the 3 boxes, the probability of drawing a red ball from this box is :

P(A) =1/4*1/3 = 1/12.

There are 3 balls in box u=2.

Among these 3 balls 1 ball is red.

The probability of drawing a red ball from box u=1 is 1/3.
Among all the 3 boxes, the probability of drawing a red ball from this box is :

P(B) =1/3*1/3 = 1/9.
There are 2 balls in box u=3.

Among these 2 balls 1 ball is red.

The probability of drawing a red ball from box u=1 is 1/2.
Among all the 3 boxes, the probability of drawing a red ball from this box is :

P(C) =1/2*1/3 = 1/6.Now the total probability P = P(A)+P(B)+P(C)

i. e. P = 1/12+1/9+1/6 = 13/36

The python code for the solution of the above problem is given below :

```
```

`s1=0`

`tries1=0`

`s2=0`

`tries2=0`

`s3=0`

`tries3=0`

`for box in range(1,4):`

` for ball in range(1,5):`

` if box==1:`

` if ball==1:`

` s1+=1`

` tries1+=1`

` for ball in range(1,4):`

` if box==2:`

` if ball==1:`

` s2+=1`

` tries2+=1`

` for ball in range(1,3):`

` if box ==3:`

` if ball==1:`

` s3+=1`

` tries3+=1`

`print((float(s1)/tries1)+(float(s2)/tries2)+(float(s3)/tries3))`

#>>> 13/36

#0.3611111111111111

#### Answer :

(b) Given that the ball is red, what are the probabilities that the chosen box was box u = 1, 2, 3?We know that box u=1 contains 4 balls among which 1 is a red ball.

The probability of drawing a red ball here is P(A)=1/4.

Similarly box u=2 contains 3 balls among which 1 is a red ball.

The probability of drawing a red ball here is P(B)=1/3.

And box u=3 contains 2 balls among which 1 is a red ball.

The probability of drawing a red ball here is P(C)=1/2.

The red ball might have been drawn from box u=1 OR u=2 OR u=3.

OR indicates addition.

Hence total probability is P=P(A)+P(B)+P(C)=1/4+1/3+1/2 = 13/12

Now the probability of drawing a red ball from box u=1 is P(A)/P = 3/13.

The probability of drawing a red ball from box u=2 is P(B)/P = 4/13.

The probability of drawing a red ball from box u=3 is P(C)/P = 6/13

The python code is given below :

```
box1=0
```

box2=0

box3=0

tries1=0

tries2=0

tries3=0

for first in range(1,5):

if first==1:

box1+=1

tries1+=1

a=float(box1)/tries1

for second in range(1,4):

if second==1:

box2+=1

tries2+=1

b=float(box2)/tries2

for third in range(1,3):

if third ==1:

box3+=1

tries3+=1

c=float(box3)/tries3

print('The probability of drawing the red ball from box u=1 is = ',float(a)/float(a+b+c))

print('The probability of drawing the red ball from box u=2 is = ',float(b)/float(a+b+c))

print('The probability of drawing the red ball from box u=3 is = ',float(c)/float(a+b+c))

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