Dogs in the GoodDog Obedience School win a blue ribbon for learning how to sit, a green ribbon for learning how to roll over, and a white ribbon for learning how to stay. There are $100$ dogs in the school. $\bullet$ $62$ have blue ribbons, $55$ have green ribbons, and $63$ have white ribbons. $\bullet$ $32$ have a blue ribbon and a green ribbon; $31$ have a green ribbon and a white ribbon; $38$ have a blue ribbon and a white ribbon. $\bullet$ $16$ have all three ribbons. How many dogs have not learned any tricks

5 dogs have not learned any tricks

Step-by-step explanation:

Let B represent the blue ribbons, G represent green ribbon and W represent white ribbon

n(B) = 62, n(G) = 55, n(W) = 63, n(B ∩ G ) = 32, n(G ∩ W ) = 31, n(B ∩ W) = 38, n(B ∩ G ∩ W) = 16

n(B ∩ G ∩ W') = those with only blue and green color = n(B ∩ G ) - n(B ∩ G ∩ W) = 32 - 16 = 16

n(B ∩ G' ∩ W) = those with only blue and white color = n(B ∩ W) - n(B ∩ G ∩ W) = 38 - 16 = 22

n(B' ∩ G ∩ W) = those with only white and green color = n(G ∩ W ) - n(B ∩ G ∩ W) = 31 - 16 = 15

n(B ∩ G' ∩ W') = those with only blue color = n(B) - n(B ∩ G ∩ W') - n(B ∩ G' ∩ W) - n(B ∩ G ∩ W) = 62 - 16 - 22 - 16 = 8

n(B' ∩ G ∩ W') = those with only green color = n(G) - n(B ∩ G ∩ W') - n(B' ∩ G ∩ W) - n(B ∩ G ∩ W) = 55 - 16 - 15 - 16 = 8

n(B' ∩ G' ∩ W) = those with only white color = n(W) - n(B' ∩ G ∩ W) - n(B ∩ G' ∩ W) - n(B ∩ G ∩ W) = 63 - 22 - 15 - 16 = 10

Those dogs without trick = n(B ∪ G ∪ W)' = 100 - n(B' ∩ G' ∩ W) - n(B' ∩ G ∩ W') - n(B ∩ G' ∩ W') - - n(B ∩ G ∩ W) - n(B ∩ G ∩ W')  - n(B' ∩ G ∩ W) - n(B ∩ G; ∩ W)  = 100 - 10 - 8 - 8 - 16 - 16 - 22 - 15 = 5

for a it would be 24 children.

for b i don't understand the question. is it saying for each individual person they pick 5 reds to 7 greens or is it the whole group. clarify and i hope this can you.

step-by-step explanation: