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# Mathcounts trainer question:

0
539
5 This is the same program as CuteDramiones, I just need help, like a nudge, then If I still don't get it, a bigger hint pls... Or if you don't want to talk, the whole solution.

Jun 23, 2019

#1
+2

P(< 15 from Box A)  = (14/20)  =   7/10      (1)

P(even or > 25 from Box B)   =

P(even from Box B) + P(>25 from Box B)  - P(even and > 25 from Box B)  =

(10/20) + (5/20) - (3/20)  = 12/20  = 3/5     (2)

So.....the probability of  (1) and (2)   =   7/10  *  3/5  =    21 /  50   Jun 23, 2019
#2
+11

Thank You, You are correct.

tommarvoloriddle  Jun 23, 2019
#3
+1

No prob.....!!!   CPhill  Jun 23, 2019
#4
+11         tommarvoloriddle  Jun 23, 2019
#5
+2

You didnt Cphills is like this Space Space See   Hear is a answer from Cphill

P(< 15 from Box A)  = (14/20)  =   7/10      (1)

P(even or > 25 from Box B)   =

P(even from Box B) + P(>25 from Box B)  - P(even and > 25 from Box B)  =

(10/20) + (5/20) - (3/20)  = 12/20  = 3/5     (2)

So.....the probability of  (1) and (2)   =   7/10  *  3/5  =    21 /  50   See I hope this helps!   Nickolas  Jun 24, 2019