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Zorluk: ZorCompound Events and Probability Laws

In an agricultural experiment, two crop varieties, XX and YY, are tested independently for germination under drought conditions. The probability that variety XX germinates is 0.650.65, and the probability that at least one of the two varieties germinates is 0.860.86. What is the probability that variety YY germinates?

Cevap: 0.6

Cevap

The probability that variety Y germinates is 0.6
Using the law of addition for independent events P(XY)=P(X)+P(Y)P(X)P(Y)P(X \cup Y) = P(X) + P(Y) - P(X)P(Y), substituting P(X)=0.65P(X) = 0.65 and P(XY)=0.86P(X \cup Y) = 0.86 gives 0.86=0.65+0.35P(Y)0.86 = 0.65 + 0.35 P(Y), which yields P(Y)=0.6P(Y) = 0.6.

Adım Adım Çözüm

1
Apply the general addition law for two probability events
P(XY)=P(X)+P(Y)P(XY)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.
2
Express the intersection using the law of multiplication for independent events
P(XY)=P(X)×P(Y)=0.65×P(Y)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.
3
Substitute given values into the combined probability formula and solve for P(Y)P(Y)
0.86=0.65+P(Y)0.65P(Y)    0.21=0.35P(Y)    P(Y)=0.60.86 = 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)P(Y) yields the correct decimal value.

Anahtar Kavram

Compound Probability Laws and Independent Events
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