Question

The probability that it will be cold enough to have snow stick to the roads on New Year;s Eve is .92. The probability that it will snow on New Years Eve, given that it is not cold enough to stick to the roads is .07.
---Draw a tree diagram and label all parts

@aum

Answer 1

were you able to draw the tree?

Answer 2

@superhelp101 Are you there?

Answer 3

yes. I was just a bit confused, it there more to it for the tree ? @kropot72

Answer 4

Sorry, I will delete my attempts and take a fresh look.

Answer 5

thx

Answer 6

@dumbcow @ganeshie8 @paki can anyone please continue?

Answer 7

\[P(A) = .92\]
\[P(A^c) = .08\]
\[P(B|A^c) = .07 \]
\[P(B and A^c) = .07*.08 = .0056\]

Answer 8

ok what is the probability that it will not snow on new years eve, given that it is cold enough for the snow to stick to the road?

Answer 9

oh wowwww! thanks guys

Answer 10

Sorry, I misunderstood the question and must delete my attempt again. :(

Answer 11

The part probability tree shows the given probabilities and the probability of being not cold enough to stick, P(not stick) which is found from:
P(not stick) = 1 - P(stick) = 1 - 0.92 = 0.08.
\[P(snow \cap not stick)=0.07=P(snow) \times P(not stick)\ ............(1)\]
From equation (1) we can find P(snow) and P(not snow) as follows:
\[P(snow)=\frac{0.07}{0.08}=0.875\]
\[P(\bar{snow})=1-0.875=0.125\]
Now the probability tree can have more values added.

Answer 12

"ok what is the probability that it will not snow on new years eve, given that it is cold enough for the snow to stick to the road?"
In this case, since the probabilities of the events 'snow' or 'not snow' are mutually exclusive we get:
\[P(snow \cup \bar{snow})=P(snow \cap stick)+P(\bar{snow} \cap stick)=0.805+0.115=0.92\]
and P(not snow) is given by:
\[\large \frac{0.115}{0.92}=0.125\]

Answer 13

thank you!!!