Answer:
[tex] \displaystyle 0.344[/tex]
Step-by-step explanation:
we are given that a coin is tossed 6 times and we want to find the probability of getting at most 2 heads.
To solve this problem,we can consider binomial distribution, which is given by
[tex] \displaystyle P(X = r) = \binom{n}{r} {p}^{r} {q}^{n - r} [/tex]
where:
- P = binomial probability
- r = number of times for a specific outcome within n trials
- [tex]{n \choose r}[/tex] = number of combinations
- p = probability of success on a single trial
- q = probability of failure on a single trial
- n = number of trials
we want to figure out the probability of getting at most 2 heads out of 6 trials , The probability can therefore be found by adding up all the binomial distributions including X=2 and less than it, Thus
[tex] \displaystyle P(X \leq 2) = P(X=0)+P(X=1)+P(X=2) [/tex]
[tex] \displaystyle P(X \leq 2) = \binom{6}{0} {p}^{0} {q}^{6 - 0} + \binom{6}{1} {p}^{1} {q}^{6- 1} + \binom{6}{2} {p}^{2} {q}^{6 - 2} [/tex]
when a coin is tossed, the probability of getting both head (success) and tail (failure) are ½ which is why ,the variables, p and q are assigned to ½. therefore substitute
[tex] \rm\displaystyle P(X \leq 2) = \binom{6}{0} { \left( \frac{1}{2} \right) }^{0} { \bigg( \frac{1}{2} \bigg) }^{ 6- 0} + \binom{6}{1} { \bigg( \frac{1}{2} \bigg) }^{1} { \bigg( \frac{1}{2} \bigg) }^{6 - 1} + \binom{6}{2} { \bigg( \frac{1}{2} \bigg)}^{2} { \bigg( \frac{1}{2} \bigg) }^{6- 2} [/tex]
since p and q are the same. it won't make any difference to write all the product of p and q as (½)⁶:
[tex]\rm\displaystyle P(X \leq 2) = \binom{6}{0} { \bigg( \frac{1}{2} \bigg) }^{ 6} + \binom{6}{1} { \bigg( \frac{1}{2} \bigg) }^{6} + \binom{6}{2} { \bigg( \frac{1}{2} \bigg) }^{6}[/tex]
In the expression the term (½)⁶ is common thus factor it out:
[tex]\rm\displaystyle P(X \leq 2) = { \bigg(\frac{1}{2}\bigg) }^{ 6} \left( \binom{6}{0} + \binom{6}{1} + \binom{6}{2} \right) [/tex]
calculate the combinations:
[tex]\rm\displaystyle P(X \leq 2) = { \bigg(\frac{1}{2}\bigg) }^{ 6} \left(1+6+15\right) [/tex]
simplify addition:
[tex]\rm\displaystyle P(X \leq 2) = { \bigg(\frac{1}{2}\bigg) }^{ 6} \left(22\right) [/tex]
simplify exponent:
[tex]\rm\displaystyle P(X \leq 2) = { \bigg(\frac{1}{64}\bigg) } \left(22\right) [/tex]
simplify multiplication:
[tex]\rm\displaystyle P(X \leq 2) = \frac{22}{64} [/tex]
dividing yields:
[tex]\rm\displaystyle P(X \leq 2) = 0.34375 [/tex]
[tex]\rm\displaystyle P(X \leq 2) \approx 0.344 [/tex]
In conclusion
The answer is 0.344
