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avatar+625 

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
avatar+101813 
+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

 

 

cool cool cool

 Jun 23, 2019
 #2
avatar+625 
+4

Thank You, You are correct.

tommarvoloriddle  Jun 23, 2019
 #3
avatar+101813 
+1

No prob.....!!!

 

cool cool cool

CPhill  Jun 23, 2019
 #4
avatar+625 
+4

coolcoolcool

coolcoolcool

coolcoolcool

 

I stole your trademark... LOL

tommarvoloriddle  Jun 23, 2019
 #5
avatar+912 
+2

You didnt Cphills is like this

 

 

coolSpacecoolSpacecool  

 

See

 

cool cool cool

 

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

 

cool cool cool

 

See I hope this helps! 

 

cool cool cool 

Nickolas  Jun 24, 2019

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