Question

Three coins are tossed together. Find the probability of getting: exactly two heads, at least two heads, at least one head and one tail

Answer 1

Step 1: Write the formula for probability and find the total number of possible outcomes:

The formula for probability is,

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Given that, three coins are tossed together.

$$\begin{gathered} n\left(S\right)={(possibleoutcomesoftos\sin gacoin)}^{numberofcoinstossed} \\ ={\left(2\right)}^3 \\ =8 \end{gathered}$$

Thus, the total number of outcomes will be $8.$

$S=\{HHH,TTT,THH,HTH,HHT,HTT,THT,TTH\}$

Step 2: Find the probability of getting exactly two heads:

Let $A$ be the event of getting exactly two heads

$\mathrm{A}=\mathrm{HHT},\mathrm{HTH},\mathrm{THH}$

$n\left(A\right)=3$

There are $3$ cases in favor.

Thus, the total outcomes $=8$

Therefore, Probability $=\frac{3}{8}$

Step 3: Find the probability of getting at least two heads:

Let $B$ be the event of getting at least two heads

$$\begin{gathered} B=HHH,HTH,THH,HHT \\ n\left(B\right)=4 \end{gathered}$$

Thus the total outcomes $=8$

$P\left(B\right)=\frac{4}{8}=\frac{1}{2}$

Step 4: Find the probability of getting at least one head and one tail:

$HTT,THT,TTH,HHT,HTH,THH=6$

There are six cases in favor

Thus the total outcomes $=8$

Therefore Probability,

$$\begin{gathered} =\frac{6}{8} \\ =\frac{3}{4} \end{gathered}$$

Hence,the probability of getting exactly two heads$=\frac{3}{8}$, getting at least two heads$=\frac{1}{2}$ and getting at least one head and one tail $=\frac{3}{4}$