Question

Four coins are tossed. Let T be the random variable representing the number of tails that occur. Construct a probability distribution table and find the probability of picking 2 tails.

Answer 1

Obviously the number of tails T we get after 4 coins are tossed may be any value of 0, 1, 2, 3, 4.

We have a Bernoulli trial - exactly two possible outcomes, "success" (we get tail) and "failure" (head) and the probability of success is the same every time the experiment is conducted (the coin is tossed).

The probability of each result:

$P(T=k)=\begin{pmatrix} n \\ k \end{pmatrix}\cdot p^k \cdot q^{n-k}=\begin{pmatrix} 4 \\ k \end{pmatrix}\cdot \begin{pmatrix} \cfrac{1}{2} \end{pmatrix}^k \cdot \begin{pmatrix} \cfrac{1}{2} \end{pmatrix}^{4-k}=$

$=\cfrac{4!}{k!\cdot(4-k)!}\cdot\begin{pmatrix} \cfrac{1}{2} \end{pmatrix}^4=\cfrac{2\cdot3\cdot4}{k!\cdot(4-k)!\cdot2^4}=\cfrac{3}{2\cdot k!\cdot(4-k)!};$

$P(T=0)=\cfrac{3}{2\cdot 0!\cdot4!}=\cfrac{3}{2\cdot1\cdot2\cdot3\cdot4}=\cfrac{1}{16};$

$P(T=1)=\cfrac{3}{2\cdot 1!\cdot3!}=\cfrac{3}{2\cdot1\cdot2\cdot3}=\cfrac{1}{4};$

$P(T=2)=\cfrac{3}{2\cdot 2!\cdot2!}=\cfrac{3}{2\cdot2\cdot2}=\cfrac{3}{8};$

$P(T=3)=\cfrac{3}{2\cdot 3!\cdot1!}=\cfrac{3}{2\cdot2\cdot3\cdot1}=\cfrac{1}{4};$

$P(T=4)=\cfrac{3}{2\cdot 4!\cdot0!}=\cfrac{3}{2\cdot2\cdot3\cdot4\cdot1}=\cfrac{1}{16}.$

The probability distribution table:

The probability of picking 2 tails:

$P(T=2)=\cfrac{3}{8}.$