Question

What is conditional probability of yy, that a pea plant has two dominant?

Answer 1

SolutionThe problem statement yields the obvious facts that P[L] = 0.16 and P[H] = 0.10.The words “10% of the ticks that had either Lyme disease or HGE carried bothdiseases” can be written asP [LH|L∪H] = 0.10.(1)(a) SinceLH⊂L∪H,P [LH|L∪H] =P [LH∩(L∪H)]P [L∪H]=P [LH]P [L∪H]= 0.10.(2)Thus,P [LH] = 0.10 P [L∪H] = 0.10 (P [L] + P [H]-P [LH]).(3)Since P[L] = 0.16 and P[H] = 0.10,P [LH] =0.10 (0.16 + 0.10)1.1= 0.0236.(4)(b) The conditional probability that a tick has HGE given that it has Lymedisease isP [H|L] =P [LH]P [L]=0.02360.16= 0.1475.(5)Problem 1.5.1 SolutionFrom the table we look to add all the mutually exclusive events to find each prob-ability.(a) The probability that a caller makes no hand-offs isP [H0] = P [LH0] + P [BH0] = 0.1 + 0.4 = 0.5.(1)(b) The probability that a call is brief isP [B] = P [BH0] + P [BH1] + P [BH2] = 0.4 + 0.1 + 0.1 = 0.6.(2)(c) The probability that a call is long or makes at least two hand-offs isP [L∪H2] = P [LH0] + P [LH1] + P [LH2] + P [BH2]= 0.1 + 0.1 + 0.2 + 0.1 = 0.5.(3)21