Two candidates, X and Y, sit for an entrance examination independently. If the probability… — JAMB UTME

Two candidates, $X$ and $Y$ , sit for an entrance examination independently. If the probability that candidate $X$ passes is $\frac{4}{7}$ and the probability that candidate $Y$ passes is $\frac{1}{3}$ , what is the probability that at least one of them passes the examination?

A
$\frac{19}{21}$
$\frac{5}{7}$ Cevap
C
$\frac{4}{21}$
D
$\frac{8}{21}$

Cevap

The probability that at least one candidate passes the examination is

\frac{5}{7}

The probability that at least one candidate passes is given by the union of independent events

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

. Substituting the given values yields

\frac{4}{7} + \frac{1}{3} - \left(\frac{4}{7} \cdot \frac{1}{3}\right) = \frac{15}{21} = \frac{5}{7}

. Alternatively, using the complement rule:

1 - P(X')P(Y') = 1 - \left(1 - \frac{4}{7}\right)\left(1 - \frac{1}{3}\right) = 1 - \left(\frac{3}{7} \cdot \frac{2}{3}\right) = 1 - \frac{2}{7} = \frac{5}{7}

Adım Adım Çözüm

Identify given probabilities and state independence condition

P(X) = \frac{4}{7}

and

P(Y) = \frac{1}{3}

. Since

X

and

Y

are independent events,

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

Independent events allow the joint probability of both events occurring to be calculated as the product of their individual probabilities.

Calculate the intersection probability

P(X \cap Y)

P(X \cap Y) = \frac{4}{7} \times \frac{1}{3} = \frac{4}{21}

The intersection gives the probability that both candidate

X

and candidate

Y

pass.

Apply the general addition law of probability

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

P(X \cup Y) = \frac{4}{7} + \frac{1}{3} - \frac{4}{21} = \frac{12}{21} + \frac{7}{21} - \frac{4}{21} = \frac{15}{21} = \frac{5}{7}

The union of two events represents the event that at least one of them occurs.

Anahtar Kavram

Probability Laws for Compound and Independent Events

Alternatif Yöntem

Using the complementary law of probability:

P(\text{at least one passes}) = 1 - P(\text{neither passes}) = 1 - P(X')P(Y') = 1 - \left(1 - \frac{4}{7}\right)\left(1 - \frac{1}{3}\right) = 1 - \left(\frac{3}{7} \times \frac{2}{3}\right) = 1 - \frac{2}{7} = \frac{5}{7}

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