Question

A purse contains 3 silver coins and 4 copper coins. Second purse

contains 4 silver and 3 copper coins. If a coin is selected at random from

one of the two purses, find the probability that it is a silver coin

(a) 19/42

(b) 1/3

(c) 2/7

(d) 5/6​

Answer 1

option -A

19/42

EXPLANATION:

Consider the following events:

E

1

=Selecting purse, E

2

=Selecting second purse, A=Coin draw is silver coin.

We have, P(E

1

)=P(E

2

)=

2

1

,P(A/E

1

)=

6

2

,P(A/E

2

)=

7

4

Required probability =P(A)=P(E

1

)P(A/E

1

)+P(E

2

)P(A/E

2

).

=

2

1

×

6

2

+

2

1

×

7

4

=

6

1

+

7

2

=

42

7+12

=

42

19

Answer 2

$\bigstar$ Given :-

Purse A contain 4 copper 3 silver

Purse B contain 6 copper 2 silver coins

Solution :-

Probability of choosing a bag = 1/2

So, From bag A probability of getting copper coin = 4/7.

And From bag B probability of getting copper coin = 6/8.

So the required probability = 1/2(4/7 + 6/8)

After solving this we get 37/56

Answer = 37/56