Question

78. Rhubarb Refer to Exercise 76. Would you be surprised if $3$ or more of the plants in the bundle die before producing any rhubarb? Calculate an appropriate probability to support your answer.

Short answer

The probability is less than $5\%$,Therefore$P(X\ge 3)=1.16\%$and the result is surprising.

Step-by-step solution

Given information

Three or more of the plants in the bundle die before producing any rhubarb.

Explanation

Given, $n=10$and $p=0.05$

The Binomial Probability:

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{c}n\\ k\end{array}\right)\times p^k\times (1-p{)}^{n-k}$

$\mathrm{P}(\mathrm{X}=0)=\left(\begin{array}{c}10\\ 0\end{array}\right)\times (0.05{)}^0\times (0.95{)}^{10-0}\approx 0.5987$

$\mathrm{P}(\mathrm{X}=1)=\left(\begin{array}{c}10\\ 1\end{array}\right)\times (0.05{)}^1\times (0.95{)}^{10-1}\approx 0.315$

$\mathrm{P}(\mathrm{X}=2)=\left(\begin{array}{c}10\\ 2\end{array}\right)\times (0.05{)}^2\times (0.95{)}^{10-2}\approx 0.0746$

$$\begin{gathered} P(X\ge 3)=1-P(X=0)-P(X=1)-P(X=2) \\ =1-0.5987-0.3151-0.0746 \\ =0.0116 \\ =1.16\mathrm{\%} \end{gathered}$$