Question

3. A bag contains 5 white and 6 red marbles. Another bag contains 4 white and
7 red marbles. Two marbles are drawn from the selected bag. What is the
probability that selected bag contains (a) white marbles.(b) one white and one
red marble.​

Answer 1

Answer:

a) Ans: 9/22

b) Ans: 1/11+1/11

= 1+1/11

=2/11

=5.5

Answer 2

Answer:

(a) probability of white marble is $\frac{9}{11}$

(b) probability of one white and one red marble is $\frac{59}{121}$

Step-by-step explanation:

Explanation :

Given, a bag contains 5 white and 6 red marbles .Another

bag contains 4 white and 7 red marbles .

Let the two bag be $I and II$.

Therefore ,

Total marble in bag $I$ = 11         (5 white marbles and 6 red marbles.)

Total marble in bag $II$ = 11        (4 white and 7 red marbles)

Part (a):

Step 1:

Total number of white  marble in bag $I$ = 5

Probability of white marbles of bag $I$   P($I$) = $\frac{5}{11}$

Similarly ,

Total number  of white marbles  in bag $II$=  4

So , the probability of white marbles of bag $I$$I$ = P($II$) = $\frac{4}{11}$

Therefore the probability of white marbles =  P($I$) + P ($II$)

⇒ $\frac{5}{11} +\frac{4}{11}$ = $\frac{9}{11}$.

Part (b):

Step1:

Let 'A' be the event that marble selected from the first bag is white and marble selected from the second bag is white.

Let 'B' be the event that marble selected from the first bag is red and marble selected from the second bag is white.

P(A) = $\frac{5}{11}$ ×$\frac{7}{11}$ = $\frac{35}{121}$  

(where $\frac{5}{11}$ is the probability of white marble for first bag and $\frac{7}{11}$is the probability of red marble of second bag )

P(A) = $\frac{6}{11}$ ×$\frac{4}{11}$ = $\frac{24}{121}$

(similarly , $\frac{6}{11}$ is the probability of red marble of first bag and $\frac{4}{11}$ is the probability of white marble of second bag .)

P = P(A) + P(B) = $\frac{35}{121}+ \frac{24}{121}$  = $\frac{59}{121}$.

Final answer :

Hence , (a) probablility of white marble is $\frac{9}{11}$ and

            (b) probability of one white and one red marble is  $\frac{59}{121}$