SAT math question?

Question

a coin with two faces is tossed three times. what is the possibility that the coin will land with the same face up all three times?

thank you so far for all the responses. im actually studying for the SAT right now, and in SparkNotes 1/8 is listed as the correct answer. i thought 1/4 tho, because as some of you said, its 1/8 + 1/8 since it can either be all heads or all tails. im pretty sure now that the book is wrong..THANK YOU TO EVERYONE for your help!!!

Answer 1

probability of heads in all three - 1/2 * 1/2 * 1/2 = 1/8
probability of tails in all three - 1/2 * 1/2 * 1/2 = 1/8

possibility that the coin will land with the same face up all three times = 1/8 + 1/8 = 1/4

Answer 2

1/8

Answer 3

One of the most useful tools for working out probabilities from 1st principles is a tree diagram.

Start at left hand side of page, about half way down and put a point to mark the start.

Nowdraw 2 branches at approx 90 degrees to each other , representing the first throw.

At the end of each branch draw another 2 twigs representing heads and tails for the second throw.

At the end of each or these draw another 2 twigs to represent the 3rd throw.

It can land

HHH,HHT,HTH,HTT,THH,THT,TTH, TTT

Of the 8 possibilities, 2 have all 3 faces the same, HHH or TTT

I make that 1/4

Answer 4

2*2*2 = 8 total possibilities. Chances for three all heads 1/8. Chances for three all tails 1/8. Chances or all the same three times(heads or tails) 2/8 = 1/4.

Answer 5

A coin can either land on Heads or Tails. For each toss, the probability it will land on a certain face (let's say heads) is 1/2, because there is 1 way it can result in heads, and 2 different ways it can result total (heads OR tails).

Since the coin is tossed three times, (1/2)*(1/2)*(1/2) = 1/8.

Answer 6

Many people will tell you it's 1/2*1/2*1/2 = 1/8, which is technically correct; however, there is an appropriate way to calculate this using binomial distribution.

[ n! / ((k!)(n-k)!) ] * (1/2)^k * (1/2)^(n-k)

n = Number of tosses
k = Desired successful outcomes

The 1/2 in (1/2)^k comes from the odds of landing the correct face up on a single toss. The 1/2 in (1/2)^(n-k) comes from the odds of landing on something else on a single toss, namely the wrong face down.

n = 3 because you specified three tosses.
k = 3 because you specified three tosses resulting in the correct face up.

P = 3!/3!0! * (1/2)^3 * (1/2)^0
P = 3!/3! * 1/8 * 1
P = 1 * 1/8 = 1/8 = 12.5%

There is a 12.5% probability that you will land the correct face up on all three tosses.

What makes this equation more useful is that it can be applied to other games of chance, including cards and dice.

eg.
The odds of landing on 1 on a six-sided dice only twice over three rolls is

n = 3
k = 2

P = 3!/2!(3-2)! * (1/6)^2 * (5/6)^1
* Notice I used 1/6 here because the odds of landing on 1 on a single roll is 1 out of 6, and the odds of landing on something else is 5 out of 6.
P = (6/2) * (1/36) * (5/6)
P = 3 * (1/36) * (5/6)
P = (1/12) * (5/6) = 5/72 = 6.9%

Answer 7

1/4. the 1st toss doesn't matter. the next 2, there is 1/2 on each toss that it will be the same as the 1st toss. 1/2 * 1/2=1/4 or 25%

Answer 8

all the three tosses are independent event

so p(head on first,second,third)=1/2*1/2*1/2=1/8

(remember in case on independent event-p(a intersection b) commonnly called as p(a and b)=p(a).p(b)

Answer 9

1/8 chances

Answer 10

Think of all of the combinations you can make with the coin. Then count how many possibilities have the same letter. Count all of the possibilities. The number of possibilities should be the denominator, and the other the numerator of the fraction. The answer is.......