A. Probability of drawing a red ball

red ball = 4

blue ball = 3

yellow ball = 3

P(red ball) =number of red balls/total number of balls` `

P(red ball) = `4/(4+3+3)`

P(red ball) = `4/10`

P(red ball) = `2/5`

B. Probability of drawing three yellow ball

Since you draw without replacement, each time you draw, the total number of is deducted and the number of yellow ball as well.

P(all three are yellow ball) =`(3/10)(2/9)(1/8)`

P(all three are yellow ball) = `6/720=1/120`

C. Probability of drawing a pair but not blue ball

The combination you can have without blue ball in drawing a pair are (R,R), (R,Y), (Y,R), (Y,Y).``

P(R,R) = `(4/10)(3/9)=12/90`

P(R,Y) = `(4/10)(3/9)=12/90`

P(Y,R) = `(3/10)(4/9)=12/90`

P(Y,Y) = `(3/10)(2/10)=6/90`

Just add the probabilities.

P(pair but not blue ball) = `42/90 =7/15`

Another solution is:

P(pair but not blue) = `(red+yellow)/sum((red+yellow)-1)/(sum-1)`

P(pair but not blue) = `((4+3)/10)(6/9)`

P(pair but not blue) = `42/90` = `7/15`