probability 2

Multiply probabilities.  P(event1) x P(event2).  0.7 x 0.3 = 0.21.

ore more than 4 goals AND next day is WET.

Please explain, unknown_routine.  Thanks.

I find no  unknown_routine
-
go with ozo       probability =    0.39

Nathan is a real good soccer player

The  0.21 is the probability that it will be wet AND 4 goals. It does not take into consideration that the day might be dry.

When it is said that the probability that it will be wet on the day of the next game
is 0.7; it means by definition that the the day may be a dry day has been taken into consideration.

That's without considering that the day may be dry, there is no need (meaning) to say that the probability that the day will be wet is 0.7.

In other words, how can there be a probability of a coin turning its head without its tail?

The question was "What is the probability of scoring four goals" not what is the probability of scoring four goals on day which may be wet, which is what you calculated.
Ozo clearly shows all the possibilities of scoring four goals on a day which may be wet but which may also be dry.
You have neglected the possibility that the day may be dry. The probability of four  goals on the possible dry day is by your type of calculation
0.6 x 0.3 = 0.18
which added to your calculation of four goals on a wet day of 0.21 gives
0.39
which is just what ozo calculated.

Why not state the whole question?: "Calculate the probability that Nathan will kick more than four goals in the next game."

[I have not commented on Ozo's post. Nor I do now.]

Why not state the whole question?

because it is not my question and because your question was not asked

The probability of your winning this dialog (with me) is 1.  Hence, I stop.   Cheers.  :-)

OK your question made me think  hard after so many years :)

I just provide you mathematical explanation of the answer, as I see other people also obtained this answer.

First we define:

p(4+)=Probabiliy that Nathan will kick more than four goals in the next
game.

P(4+ AND W)= Probabiliy that Nathan will kick more than four goals in the next
game AND next day is Wet.

P(4+ AND D)= Probabiliy that Nathan will kick more than four goals in the next
game AND next day is Dry.

P(4+|W)=Probabiliy that Nathan will kick more than four goals in the next
game IF next day is Wet=0.3 (according to the question)
 
P(4+|D)=Probabiliy that Nathan will kick more than four goals in the next
game IF next day is Dry=0.6 (according to the question)

P(W)=Probability of next day of the next game is WET= 0.7(according to the question)

P(D)=Probability of next day of the next game is Dry= 1-0.7=0.3

By the definiton of the "Conditional probability":

P(4+ AND W)= P(4+|W) * P(W)=0.3 *0.7=0.21

P(4+ AND D)= P(4+|D) * P(D)=0.6*0.3=0.18

Therefore: probability that Nathan will kick more than four goals in the next
game:

P(4+)= P(4+ AND W) + P(4+ AND D) =  0.21 + 0.18=0.39

Thanks,  unknown_routine. :-)

That is the correct answer, clearly explained. :-)

And thanks, Ozo.  :-)

Another technique for solving these sort of problems quickly is to draw the
full probability tree or sample space.
ExEx-Probability.png

[d]-glitch:  Excellent!  :-)

Much thanks. :-)