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MadDaddy95

23.08.2020

Mathematics

$\begin{array}{cccccc}\text{Number of milk chocolate bars}&0&1&2&3&4\\ \\\text{Probabilities}&0.0256&0.1536&0.3456&0.3456&0.1296\end{array}$

Step-by-step explanation:

There are 10 chocolate bars in the box, 6 of them are milk chocolate bars. So the probability of chosing milk chocolate bar is $p=\frac{6}{10}=0.6$ and the probability of chosing not milk chocolate bar is $q=\frac{10-6}{10}=0.4.$

Chosing 4 chocolate bars, you can choose 0, 1, 2, 3, 4 milk chocolate bars. Find probabilities:

$P(\text{Number of milk chocolate bars}=0)=C^{4}_0p^0q^4=1\cdot 0.6^0\cdot 0.4^4=0.4^4=0.0256$

$P(\text{Number of milk chocolate bars}=1)=C^{4}_1p^1q^3=4\cdot 0.6^1\cdot 0.4^3=4\cdot 0.6\cdot 0.064=0.1536$

$P(\text{Number of milk chocolate bars}=2)=C^{4}_2p^2q^2=6\cdot 0.6^2\cdot 0.4^2=6\cdot 0.36\cdot 0.16=0.3456$

$P(\text{Number of milk chocolate bars}=3)=C^{4}_3p^3q^1=4\cdot 0.6^3\cdot 0.4^1=4\cdot 0.216\cdot 0.4=0.3456$

$P(\text{Number of milk chocolate bars}=4)=C^{4}_4p^4q^0=1\cdot 0.6^4\cdot 0.4^0=1\cdot 0.1296\cdot 1=0.1296$

So, a probability distribution table is

$\begin{array}{cccccc}\text{Number of milk chocolate bars}&0&1&2&3&4\\ \\\text{Probabilities}&0.0256&0.1536&0.3456&0.3456&0.1296\end{array}$

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