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John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10.

What is the probability that he either hits a Bull’s eye or scores 10 points?

A1/3
B2/3
C3/5
D2/5
E1/4

Verified Answer :

The correct answer for the question that is being presented above is this one: "d. 2/5." John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10. The probability that he either hits a Bull's eye or scores 10 points is 2/5

Answer: The correct option is (d). \(\dfrac{2}{5}.\)

Step-by-step explanation: Given that John is playing a game of darts. The probability that he throws a dart into the centre of the dart board (the Bull’s eye) is \(\dfrac{1}{10}\) and the probability that he throws the dart into the 10-point ring is \(\dfrac{3}{10}.\)

We are to find the probability that he either hits a Bull’s eye or scores 10 points.

Let, 'A' and 'B' represents the events that John throws the dart into a Bull's eye and 10-point ring respectively.

Then, according to the given information, we have

\(P(A)=\dfrac{1}{10},~~P(B)=\dfrac{3}{10},~~~P(A\cup B)=?\)

Since John cannot throw the dart into the Bull's eye and 10 point ring together, both the events are independent of each other.

Therefore,

\(P(A\cap B)=0\)

From the theorems of probability, we have

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac{1}{10}+\dfrac{3}{10}-0=\dfrac{4}{10}=\dfrac{2}{5}.\)

Therefore, the probability that John either hits a Bull’s eye or scores 10 points is \(\dfrac{2}{5}.\)

Thus, (d) is the correct option.

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