Respuesta :
Solution:
In the toss of a coin, the possible outcome is either head (H) or tail (T).
[tex]\begin{gathered} \text{Number of heads (H) = 1} \\ \text{Number of tails (T) = 1} \\ \text{Total outcome = 2} \end{gathered}[/tex][tex]\begin{gathered} \text{Probability}=\frac{\text{number of required outcome}}{\text{total possible outcome}} \\ \\ \text{Probability of head, P(H) =}\frac{1}{2} \\ \text{Probability of tail, P(T) =}\frac{1}{2} \\ \end{gathered}[/tex]To get between 3 and 6 heads, inclusive, the following outcomes holds;
[tex]\begin{gathered} 3\text{ heads, 7 tails = 3H, 7T}=\text{HHHTTTTTTT} \\ 4\text{ heads, 6 tails = 4H, 6T}=\text{HHHHTTTTTT} \\ 5\text{ heads, 5 tails = 5H, 5T}=\text{HHHHHTTTTT} \\ 6\text{ heads, 4 tails = 6H, 4T}=\text{HHHHHHTTTT} \end{gathered}[/tex]Probability of 3H, 7T using binomial distribution for the arrangement of the outcomes
[tex]\begin{gathered} P(\text{HHHTTTTTTT)}=^{10}C_3(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}) \\ P(\text{HHHTTTTTTT)=120}\times(\frac{1}{2})^{10} \\ P(\text{HHHTTTTTTT)}=120\times\frac{1}{1024} \\ P(\text{HHHTTTTTTT)}=\frac{120}{1024} \end{gathered}[/tex]Probability of 4H, 6T using binomial distribution for the arrangement of the outcomes
[tex]\begin{gathered} P(\text{HHHHTTTTTT)}=^{10}C_4(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}) \\ P(\text{HHHHTTTTTT)=210}\times(\frac{1}{2})^{10} \\ P(\text{HHHHTTTTTT)}=210\times\frac{1}{1024} \\ P(\text{HHHHTTTTTT)}=\frac{210}{1024} \end{gathered}[/tex]Probability of 5H, 5T using binomial distribution for the arrangement of the outcomes
[tex]\begin{gathered} P(\text{HHHHHTTTTT)}=^{10}C_5(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}) \\ P(\text{HHHHHTTTTT)=252}\times(\frac{1}{2})^{10} \\ P(\text{HHHHHTTTTT)}=252\times\frac{1}{1024} \\ P(\text{HHHHHTTTTT)}=\frac{252}{1024} \end{gathered}[/tex]Probability of 6H, 4T using binomial distribution for the arrangement of the outcomes
[tex]\begin{gathered} P(\text{HHHHHTTTTT)}=^{10}C_6(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}) \\ P(\text{HHHHHTTTTT)=210}\times(\frac{1}{2})^{10} \\ P(\text{HHHHHTTTTT)}=210\times\frac{1}{1024} \\ P(\text{HHHHHTTTTT)}=\frac{210}{1024} \end{gathered}[/tex]The probability of getting between 3 and 6 heads inclusive in 10 tosses of a coin will be the sum of the probabilities gotten.
[tex]\begin{gathered} P(3\text{ and 6 heads inclusive)=}\frac{120}{1024}+\frac{210}{1024}+\frac{252}{1024}+\frac{210}{1024} \\ =\frac{792}{1024} \\ =\frac{99}{128} \end{gathered}[/tex]The probability of getting between 3 and 6 heads inclusive in 10 tosses of a coin is
[tex]\frac{99}{128}[/tex]