Question

An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace them. a) State the general ways in which the person could get a red marble and a blue marble b) State the number of ways this can occur. c) What is the probability the person gets a red and a blue marble? P(R & B) =

Answer 1

Answer with explanation:

Given : An urn contains 2 red marbles and 3 blue marbles.

Total marbles = 2+3=5

a) The general ways in which the person could get a red marble and a blue marble are :

1)  He draws red marble first and then second marble as blue.

2)  He draws blue marble first marble and then second marble as red.

b)  The number of ways to get one red and one blue marble is given by :-

$^2C_1\times^3C_1=2\times3=6$                          (i)

c) Number of ways to get 2 marbles from 5 is given by :-

$^5C_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5\times4\times3!}{3!\times2}=10$     (ii)

Now, The probability the person gets a red and a blue marble will be :-

$P(R\ \&B)=\dfrac{6}{10}=0.6$       [Divide (i) by (ii)]

Hence, the  probability the person gets a red and a blue marble= 0.6