MHB IGCSE Probability Question [ either / or ]

[shoshuban]

Captain Ahab sails The Laertes along the south coast of Madeira. On any day, the probability of him seeing a dolphin is 9/10 , the probability of him seeing a whale is 2/3 and the probability of him seeing a turtle is 1/5.
Q. Calculate the probability that on any day Captain Ahab will see either a dolphin or a whale.

The answer, according to the mark scheme, is 29/30.

Weirdly, if we add the three probabilities, it shows a value more than 1 ! ( 9/10 + 2/3 + 1/5 = 53/30 ! )

I have tried doing this. Trial 1 :
P( dolphin or whale) = 9/10 + 2/3 = 47/30, which is obviously not correct.

Trial 2:
P [(dolphin and no whale) or (whale and no dolphin)]
= (9/10 x 1/3) + ( 2/3 x 1/10 ) = 11/30 ( which doesn't agree with the mark scheme )

The methods shown in the mark scheme is :
Method 1 -> ( 9/10 x 1/3 ) + 2/3 = 29/30.
Method 2 -> 1-(1/10 x 1/3) = 29/30

Please explain me what's happening here.
1) Why is the total probability more than one in the first place
2) how do the mark scheme methods make sense?
3) What was I doing wrong in my Trial 2?

 

[MarkFL]

To explain why the 2 probabilities simply added is more than 1, let's look at a Venn Diagram:

View attachment 5014

The probability we want is the sum of the 3 colored areas. If we add the probabilities represented by the two circles, we are adding the green area twice, which represents the probability of seeing both a whale and a dolphin, so we need to subtract it out:

$$P(X)=\frac{9}{10}+\frac{2}{3}-\frac{9}{10}\cdot\frac{2}{3}=\frac{9}{10}\left(1-\frac{2}{3}\right)+\frac{2}{3}=\frac{9}{10}\cdot\frac{1}{3}+\frac{2}{3}=\frac{29}{30}$$

This was the first method used on the marking scheme.

The second method relies on the fact that it is certain either a dolphin or a whale will be spotted OR neither will be spotted, which we may state as:

$$P(X)+\frac{1}{10}\cdot\frac{1}{3}=1$$

Solving for $P(X)$, we then obtain:

$$P(X)=\frac{29}{30}$$