Probability

An experiment was conducted consisting of crossing white and purple peas, assuming that the experimental plants had not been previously crossed. According to the rules of inheritance, it can be expected that 3/4 of the new offspring will bloom purple and 1/4 white. 10 plants have germinated. Determine the probability that at least three plants will bloom purple.

Final Answer:

p =  99.8512 %

Step-by-step explanation:

$$\begin{gathered} C_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ c=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=120 \\ a=3/4=\frac{3}{4}=0.75 \\ b=1-a=1-0.75=\frac{1}{4}=0.25 \\ {p}_0=c\cdot a^0\cdot b^{10-0}=120\cdot 0.75^0\cdot 0.25^{10-0}\doteq 0.0001 \\ {p}_1=c\cdot a^1\cdot b^{10-1}=120\cdot 0.75^1\cdot 0.25^{10-1}\doteq 0.0003 \\ {p}_2=c\cdot a^2\cdot b^{10-2}=120\cdot 0.75^2\cdot 0.25^{10-2}\doteq 0.001 \\ p=100-100\cdot (p_0+p_1+p_2)=100-100\cdot (0.0001+0.0003+0.001)=99.8512\% \end{gathered}$$

Try another example