Question

A jar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10. A marble is drawn at random from the jar. Find the probability the marble is blue or evennumbered.\

Answer 1

Answer:

Step-by-step explanation:

As per given ,  The jar contains

Red marbles = (R1) , (R2) , (R3) , (R4)

Blue marbles =  (B1) ,   (B2) ,   (B3) ,   (B4) ,   (B5) ,   (B6) ,   (B7) ,   (B8) ,   (B9) ,   (B10) .

Total marbles = 4+10=14

The marbles that has even number = (R2)  , (R4) ,(B2) , (B4) ,  (B6) ,   (B8) , (B10)

=7

Total Blue marbles  = 10

Blue and even marbles = 5

Now , the number of marbles are blue or even numbered :

n(Blue or even )= n(Blue) + n(even)- n(Blue and even)

= 10+7-5 =12

Now , the probability the marble is blue or even numbered will be :

$P(\text{Blue or even }) = \dfrac{n(\text{Blue or even })}{\text{Total marbles}}\\\\=\dfrac{12}{14}=\dfrac{6}{7}$

Hence, required probability =$\dfrac{6}{7}$

Answer 2

we are given

A jar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10

so, total number of marbles =10+4=14

the number of  marble that  is red or odd number=2

so, the probability of the marble is red or odd number is

$=\frac{2}{14}$

the probability the marble is blue or even numbered

= 1- P( the marble is red or odd number)

so, we get

the probability the marble is blue or even numbered  is

$=1-\frac{2}{14}$

$=\frac{12}{14}$

$=\frac{6}{7}$...............Answer