In an agricultural trial, two independent seed varieties, V_1 and V_2, have germination pr… — JAMB UTME

In an agricultural trial, two independent seed varieties, $V_1$ and $V_2$ , have germination probabilities $P(V_1) = p$ and $P(V_2) = p + 0.20$ . If the probability that at least one seed variety germinates is $0.76$ , what is the probability that only variety $V_2$ germinates?

0.36Answer
B
0.48
C
0.52
D
0.24

Answer

The probability that only variety

V_2

germinates is

0.36

Using the addition law

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1)P(V_2)

gives

0.76 = 2p + 0.20 - (p^2 + 0.20p)

. Solving

p^2 - 1.80p + 0.56 = 0

gives

p = 0.40

. Thus

P(V_1) = 0.40

and

P(V_2) = 0.60

. The probability that only variety

V_2

germinates is

P(V_2) \cdot P(V_1') = 0.60 \times 0.60 = 0.36

Step-by-Step Solution

Set up the probability addition law for independent events.

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1 \cap V_2)

. Since

V_1

and

V_2

are independent,

P(V_1 \cap V_2) = P(V_1) \cdot P(V_2) = p(p + 0.20)

Independent events allow the intersection probability to be expressed as the product of their individual probabilities.

Substitute the given values into the addition law and solve for

p

0.76 = p + (p + 0.20) - p(p + 0.20) \implies p^2 - 1.80p + 0.56 = 0 \implies (p - 0.40)(p - 1.40) = 0

. Since

p \le 1

p = 0.40

Formulating a quadratic equation yields the value of

p

within valid probability bounds.

Calculate individual probabilities

P(V_1)

and

P(V_2)

P(V_1) = 0.40

and

P(V_2) = 0.40 + 0.20 = 0.60

Knowing

p

gives the exact germination probabilities for both varieties.

Find the probability that only variety

V_2

germinates.

P(\text{only } V_2) = P(V_2 \cap V_1') = P(V_2) \times [1 - P(V_1)] = 0.60 \times (1 - 0.40) = 0.60 \times 0.60 = 0.36

Only

V_2

germinating means

V_2

germinates and

V_1

fails to germinate.

Key Concept

Probability laws for independent compound events

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