Respuesta :
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
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)
[TeX]= (\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} )[/TeX]=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) [TeX]= \frac{20}{50} X \frac{19}{49} X \frac{18}{48} [/TeX]=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
