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Question: There are two small boxes A and B. In A there are 9 white beads and 8 black beads. In B there are 7 ...

There are two small boxes A and B. In A there are 9 white beads and 8 black beads. In B there are 7 white and 8 black beads. We want to take a bead from the box.
(A) What is the probability of getting a white bead from each box?
(B) A white bead and a black bead are added to box B and then a bead is taken from it.
What is the probability of getting a white bead from it?

Explanation

Solution

Hint: In this question, we have to find the probability of some events given. Probability of an event is given by P(E)=Number of favourable outcomesTotal number of outcomesP\left( E \right) = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}

Complete step-by-step answer:
Total number of beads in box A = 9 white + 8 black = 17
Total number of beads in box B = 7 white + 8 black = 15

(A)
Let E1={E_1} = Event of getting white bead from box A
E2={E_2} = Event of getting white bead from box B.
Probability for E1=P(E1)=Total number of whitebeadsTotal beads=917{E_1} = P({E_1}) = \dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}whitebeads}}{{Total{\text{ }}beads}} = \dfrac{9}{{17}}
Probability for E2=P(E2)=715{E_2} = P({E_2}) = \dfrac{7}{{15}}

(B)
Now, a white bead and a black bead are added to box B.
Total number of white beads in box B = 7+1=8 white
Total number of black beads in box B = 8+1 = 9 black
Total number of beads in box B = 8 white + 9 black = 17
Now, probability of E2=P(E2)=817{E_2} = P({E_2}) = \dfrac{8}{{17}}

Note: Probability of any event is the ratio of number of favorable outcomes of the event to the total number of the outcomes. Probability of any event lies always between 0 and 1