Question

If you were flipping a fair coin, what is the probability of the following happening:a.Flipping it two times and getting 1 head and 1 tail (in any order)b.Flipping it three times and getting exactly 2 heads. c.Flipping it once and getting a tail if you’ve already flipped it five times and gotten 5 tails in a row

Answer 1

Answer:

$\begin{gathered} P(a)=\frac{1}{2}=\text{50}\operatorname{\%} \\ P(b)=\frac{1}{2}=\text{ 50\%} \\ P(c)=\frac{1}{2}=\text{50}\operatorname{\%} \end{gathered}$

Step-by-step explanation:

The probability is represented by the following equation:

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

a) The probability of n tosses going 1 H and 1 Tail in 2 times is:

$\begin{gathered} P=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ P=\text{ 50\%} \end{gathered}$

b) for flipping it three times and getting exactly two heads:

Sample: {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT} Therefore, the total number of possible outcomes is 8.

$\begin{gathered} P(B)=P(\text{ getting two heads\rparen+ P\lparen getting 3 heads\rparen} \\ P(B)=\frac{3}{8}+\frac{1}{8} \\ P(B)=\frac{4}{8}=\frac{1}{2}=\text{ 50\%} \end{gathered}$

c. For flipping it once and getting a tail if you've already flipped it five times and gotten 5 tails in a row.

The prior flipping does not affect the result of the new toss, they are not dependent events.

$P(c)=\frac{1}{2}=\text{ 50\%}$