STA256 Lecture 05

• STA256H5F_Fall2025_Ch1_Probability_Distributions.pdf

• Exercise: Maize Seeds

  • Answers:
    • 1: 0.116
    • 2: 0.728
  • Reference
  • 40% percent of all seeds carry single spikelets, while 60% carry paired spikelets. A seed with single spikelets will produce maize ears with single spikelets 29% of the time and paired spikelets 71% of the time. A seed with paired spikelets will produce maize ears with single spikelets 26% of the time and paired spikelets 74% of the time
  • $P(SS)=0.4$
  • $P(PS)=0.6$
  • $P(E_S|SS)=0.29$
  • $P(E_P|SS)=0.71$
  • $P(E_S|PS)=0.26$
  • $P(E_P|PS)=0.74$
  • Reference
  • Find the probability that a randomly selected maize ear seed carries a single spikelet and produces ears with single spikelets.
    • $P(SS\cap E_S)=P(SS)P(E_S|SS)$
    • $P(SS\cap E_S)=0.4(0.29)$
    • $P(SS\cap E_S)=0.116$
  • Reference
  • Find the probability that a randomly selected maize ear seed produces ears with paired spikelets.
    • Because $P(SS)$ and $P(PS)$ are disjoint and they partition the space. We can use the Law of Total Probability.
    • $P(E_P)=P(E_P|SS)P(SS)+P(E_P|PS)P(PS)$
    • $P(E_P)=0.71(0.4)+0.74(0.6)$
    • $P(E_P)=0.728$

• Statistical Independence

• Mutually exclusive events are never independent

  • $P(A)>0,P(B)>0$
  • If they're exclusive, then $P(A\cap B)=0$
  • so $P(A|B)=\frac{P(A\cap B)}{P(B)}$
  • But $P(A\cap B)=0$
  • $P(A\cap B)=P(A)\cdot P(B)$
  • There are no $(x,y)\in {\mathbb{R}}^{+}:xy=0$

• Random Variables

• Probability Distribution Function

• Probability Mass Function

• Cumulative Distribution Function