Question

Suppose a class of 50 students has 20 males and 30 females. The insturctor will pick 1 (different) student to compete in 3 different national competitions--in mathematics, chemistry, and English.a. What is the probability that there is exactly 1 female student selected?b. What is the probability that there are at least 2 male students selected?

Answer 1

Answer:

(a)0.2908

(b)0.349

Step-by-step explanation:

Total Number of Students=50

Number of Male Students=20

Number of Female Students=30

Note that the selection is without replacement as each student selected is different. Also the student could be picked for any of Mathematics, Chemistry or English competitions in that order.

(a)Probability of Picking exactly one female student.

P(Exactly one Female)= P(FMM) OR P(MFM) OR P(MMF)

Where P(FMM) means a female is selected for Mathematics, Male for Chemistry and Male for English respectively.

P(FMM) OR P(MFM) OR P(MMF)

$= (\frac{30}{50} X \frac{20}{49} X \frac{19}{48} ) + (\frac{20}{50} X \frac{30}{49} X \frac{19}{48} ) + (\frac{20}{50} X \frac{19}{49} X \frac{30}{48} )$=0.2908

(b)Probability that at least 2 male students are selected.

P(at least 2 male students are selected) = P(FMM) OR P(MFM) OR P(MMF) OR P(MMM)

P(MMM) $= \frac{20}{50} X \frac{19}{49} X \frac{18}{48}$=0.0582

Recall from Part (a) that:

P(FMM) OR P(MFM) OR P(MMF) =0.2908

Therefore:

P(at least 2 male students are selected) =0.2908+0.0582 =0.349

Answer 2

Answer:

Step-by-step explanation:

Total = 50

Males = 20

Females = 30

a) P(exactly 1 female) = P(1 female , 2 male)

(30C1 * 20C2) / (50C3) = 30*190/19,600 = 0.2908

b) P(at least 2 males) = 1 - (P(0 males) + P(1 males))

= 1 - 30C3 * 20C0/50C3 - 30C2 * 20C1/50C3

= 1 - 4060*1/19,600 - 435*20/19,600 = 6840/19600 = 0.3489