Of all the trees planted by a landscaping firm, 45% survive. What is the probability that 13 or more of the 15 trees they just planted will survive

ANSWER:  The probability for 13 or more of the 15 trees they just planted will survive is 0.00110702414 SOLUTION: Given, The total number of plants is, n=15 The chance of survival of trees is, p=45%.  Probability of getting success p(success) = [tex]\frac{n o \text { of favourable outcomes }}{\text {total possible outcomes}}[/tex]P(success) = [tex]\frac{45}{100}[/tex]P(success) = 0.45Binomial distribution formula is given as[tex]\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{n}_{\mathrm{C} \mathrm{x}}(\mathrm{p})^{\mathrm{x}} \cdot(1-\mathrm{p})^{\mathrm{n}-\mathrm{x}}[/tex]In our case, x is greater than or equal to 13, i.e. x [tex]\geq[/tex] 13The probability for 13 or more of the 15 trees they just planted will survive is given by[tex]\mathrm{P}(\mathrm{X} \geq 13)=\mathrm{P}(\mathrm{X}=13)+\mathrm{P}(\mathrm{X}=14)+\mathrm{P}(\mathrm{X}=15)[/tex][tex]\mathrm{P}(\mathrm{X} \geq 13)=\left(15 \mathrm{C}_{13} \times(0.45)^{13} \times(1-0.45)^{15-13}\right.[/tex] + [tex]\left(^{15} \mathrm{C}_{14} \times(0.45)^{14} \times(1-0.45)^{15-14}\right.[/tex] + [tex]\left(15 \mathrm{C}_{15} \times(0.45)^{15} \times(1-0.45)^{15-15}\right.[/tex]on simplification we get[tex]=(15 \times 7) \times(0.45)^{13} \times(0.55)^{2}+15 \times(0.45)^{14} \times(0.55)^{1}+1 \times(0.45)^{15} \times 1[/tex][tex]\begin{array}{c}{=(105 \times 0.00003102863 \times 0.3025)+(15 \times 0.00001396288 \times 0.55)+} \\ {0.00000628329}\end{array}[/tex] =0.00098554703 + 0.0001151938 + 0.00000628329 = 0.00110702414Hence, the probability for 13 or more of the 15 trees they just planted will survive is 0.00110702414