Let (2, F. P) be a probability space and let E, and Ey be two events with P(E₁) = 0.2. P(E₂) 0.4 and P(E₁ E₂)=0.1.
Find the probability that (a) Exactly one of the events E, or E₂ will occur.
(b) At least one of the events E, or E, will occur.
(c) None of E, and Es will occur.
Solution:
P(E₁) = 0.2. P(E₂) 0.4 and P(E₁"\\cap" E₂)=0.1.
(a) Exactly one of the events E, or E₂ will occur.
Required probability = P(E₁) + P(E₂) - 2P(E₁"\\cap" E₂)
"=0.2+0.4-2\\times0.1\n\\\\=0.6-0.2\n\\\\=0.4"
(b) At least one of the events E1 or E2 will occur.
Required probability = P(E₁"\\cup" E₂) = P(E₁) + P(E₂) - P(E₁"\\cap" E₂)
"=0.2+0.4-0.1\n\\\\=0.6-0.1\n\\\\=0.5"
(c) None of E1 and E2 will occur.
Required probability = P(E1' "\\cap"E2') = P[(E₁"\\cup" E₂)'] = 1-P[(E₁"\\cup" E₂)]
"=1-0.5\n\\\\=0.5"
Comments
Leave a comment