Question

If a fair coin is tossed 6 times, what is the probability, rounded to the nearest thousandth, of getting at most 2 heads?​

Answer 1

`\displaystyle 0.344`

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

`\displaystyle P(X = r) = \binom{n}{r} {p}^(r) {q}^(n - r)`

where:

• P = binomial probability
• r = number of times for a specific outcome within n trials
• 
  `{n \choose r}` = 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

`\displaystyle P(X \leq 2) = P(X=0)+P(X=1)+P(X=2)`

`\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)`

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

`\rm\displaystyle P(X \leq 2) = \binom{6}{0} { \left( (1)/(2) \right) }^(0) { \bigg( (1)/(2) \bigg) }^( 6- 0) + \binom{6}{1} { \bigg( (1)/(2) \bigg) }^(1) { \bigg( (1)/(2) \bigg) }^(6 - 1) + \binom{6}{2} { \bigg( (1)/(2) \bigg)}^(2) { \bigg( (1)/(2) \bigg) }^(6- 2)`

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:

`\rm\displaystyle P(X \leq 2) = \binom{6}{0} { \bigg( (1)/(2) \bigg) }^( 6) + \binom{6}{1} { \bigg( (1)/(2) \bigg) }^(6) + \binom{6}{2} { \bigg( (1)/(2) \bigg) }^(6)`

In the expression the term (½)⁶ is common thus factor it out:

`\rm\displaystyle P(X \leq 2) = { \bigg((1)/(2)\bigg) }^( 6) \left( \binom{6}{0} + \binom{6}{1} + \binom{6}{2} \right)`

calculate the combinations:

`\rm\displaystyle P(X \leq 2) = { \bigg((1)/(2)\bigg) }^( 6) \left(1+6+15\right)`

simplify addition:

`\rm\displaystyle P(X \leq 2) = { \bigg((1)/(2)\bigg) }^( 6) \left(22\right)`

simplify exponent:

`\rm\displaystyle P(X \leq 2) = { \bigg((1)/(64)\bigg) } \left(22\right)`

simplify multiplication:

`\rm\displaystyle P(X \leq 2) = (22)/(64)`

dividing yields:

`\rm\displaystyle P(X \leq 2) = 0.34375`

`\rm\displaystyle P(X \leq 2) \approx 0.344`

In conclusion

The answer is 0.344