Ella's college professors assign homework independently from each other. Ella knows that there is a 60% chance that she will have math homework tonight and a 70% chance that she will have English homework. What is the probability that she will NOT have math or English homework tonight?

Question

ganeshie8 · Answer

So, she does not want any home work tonight. that means you need to find  :

P("no math homework" AND "no english homework" )

ganeshie8 · Answer

since both events are independent, you can simply multiply the probabilities

ganeshie8 · Answer

P("no math homework" AND "no english homework" ) =  P("no math homework") * P("english homework")

anonymous · Answer

42%?

ganeshie8 · Answer

nope, try again

anonymous · Answer

That is what I got when I multiplied.

ganeshie8 · Answer

you need to find P("no math homework")  first

ganeshie8 · Answer

P("no math homework")  =  1-  P("getting math homework"
                                    =  ?

anonymous · Answer

I need to know how, I don't expect you to give me a hand out, but help?

ganeshie8 · Answer

sure :)

ganeshie8 · Answer

P("no math homework")  =  1-  P("getting math homework")
                                    =  1 - 0.6
                                    = 0.4

ganeshie8 · Answer

similarly,

P("no english homework")  =  1-  P("getting english homework")
                                    =  1 - 0.7
                                    = 0.3

ganeshie8 · Answer

therefore,

P("no math homework" AND "no english homework" ) =  0.4 * 0.3

ganeshie8 · Answer

that gives u 12%

ganeshie8 · Answer

let me knw if smthng doesnt make sense

anonymous · Answer

That other one I got wrong and then got one just like it, Carson drives to school the same way each day and there are two independent traffic lights on his trip to school. He knows that there is a 30% chance that he will have to stop at the first light and an 80% chance that he will have to stop at the second light. What is the probability that he will NOT have to stop at either light?    Do you simply turn them into decimals and then multiply?

anonymous · Answer

I got 14% on this one.

ganeshie8 · Answer

probability for "stopping at first light" =  30%

that means, probability for "NOT stopping at first light" = $70\%$

ganeshie8 · Answer

probability for "stopping at second light" =  80%

that means, probability for "NOT stopping at second light" = $20\%$

ganeshie8 · Answer

So, "probability for NOT stopping at either lights" =  0.7 * 0.2 =  0.14 = $ 14\% $

ganeshie8 · Answer

you're right !!

anonymous · Answer

Thank you so much for your time and actually helping me understand! I appreciate that so much. :)

ganeshie8 · Answer

np... u wlc :)

anonymous · Answer

Yeah I got 12% too and it was correct :)