If Benjie throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?
In tossing a coin, what is the probability of getting three consecutive heads or tails
New forum topics
- strength of materials: normal stress of hollow circular tube
- Please help me solve this problem: Moment capacity of a rectangular timber beam
- Solid Mensuration: Prismatoid
- Differential Equation: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0
- Differential Equation: y' = x^3 - 2xy, where y(1)=1 and y' = 2(2x-y) that passes through (0,1)
- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates: Question for Problem #12
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
We can use the geometric distribution to get the probability of an event (success) occurring after a number of failures.
If repeated independent trials can result in a success with probability $p$ and a failure with probability $q = 1 - p$, then the probability distribution of the random variable $X$, the number of trials on which the first success occurs, is:
$$g(x;p) = pq^{x-1}$$
and $x = 1,2,3,4,5,.....$
In this context, the first success is getting three consecutive heads or three consecutive tails in flipping coins. The probability of getting a head (or tail) in coin toss is $p = 0.5$. The probability that the first success occurs after flipping the coin four times would be:
$$g(x;p) = pq^{x-1}$$ $$g(4;0.5) = 0.5(1-0.5)^{(4)-1} = 0.0625 $$
Therefore, the probability that the game will end on the fourth throw is $\color{green}{6.25 \%}$
Alternate solutions are encouraged....