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
[tex]=\frac{2}{14}[/tex]
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
[tex]=1-\frac{2}{14}[/tex]
[tex]=\frac{12}{14}[/tex]
[tex]=\frac{6}{7}[/tex]...............Answer
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 :
[tex]P(\text{Blue or even }) = \dfrac{n(\text{Blue or even })}{\text{Total marbles}}\\\\=\dfrac{12}{14}=\dfrac{6}{7}[/tex]
Hence, required probability =[tex]\dfrac{6}{7}[/tex]
