In an agricultural experiment, two crop varieties, X and Y, are tested independently for g… — JAMB UTME

In an agricultural experiment, two crop varieties, $X$ and $Y$ , are tested independently for germination under drought conditions. The probability that variety $X$ germinates is $0.65$ , and the probability that at least one of the two varieties germinates is $0.86$ . What is the probability that variety $Y$ germinates?

Answer: 0.6

Answer

The probability that variety Y germinates is 0.6

Using the law of addition for independent events

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)

, substituting

P(X) = 0.65

and

P(X \cup Y) = 0.86

gives

0.86 = 0.65 + 0.35 P(Y)

, which yields

P(Y) = 0.6

Step-by-Step Solution

Apply the general addition law for two probability events

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)

The addition law relates the union, individual probabilities, and intersection of compound events.

Express the intersection using the law of multiplication for independent events

P(X \cap Y) = P(X) \times P(Y) = 0.65 \times P(Y)

Because germination of variety X and variety Y are independent events.

Substitute given values into the combined probability formula and solve for

P(Y)

0.86 = 0.65 + P(Y) - 0.65 P(Y) \implies 0.21 = 0.35 P(Y) \implies P(Y) = 0.6

Isolating the unknown probability

P(Y)

yields the correct decimal value.

Key Concept

Compound Probability Laws and Independent Events

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