Three cartridges - shooting

A shooter has three cartridges. He decided that he would shoot at the target until he hits for the first time. The probability of a hit at each shot is 0.6. The random variable X gives the number of cartridges fired.
a) Write the probability distribution of the random variable X and its distribution function.
b) Determine the mode of this random variable?
c) What is the probability that the shooter fires at most twice?

Final Answer:

b =  1
c =  0.84

Step-by-step explanation:

$$\begin{gathered} p=0.6 \\ {p}_1=p=0.6=\frac{3}{5} \\ {p}_2=(1-p)\cdot p=(1-0.6)\cdot 0.6=\frac{6}{25}=0.24 \\ {p}_3=(1-p)\cdot (1-p)\cdot p+(1-p)\cdot (1-p)\cdot (1-p)=(1-0.6)\cdot (1-0.6)\cdot 0.6+(1-0.6)\cdot (1-0.6)\cdot (1-0.6)=\frac{4}{25}=0.16 \\ distibution: \\ {F}_1=p=0.6=\frac{3}{5} \\ {F}_2=p_1+p_2=0.6+0.24=\frac{21}{25}=0.84 \\ {F}_3=p_1+p_2+p_3=0.6+0.24+0.16=1 \\ b=1 \end{gathered}$$

$c=p_1+p_2=0.6+0.24=0.84$

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