We use the equation for repeated trials written below:
Probability = n!/r!(n-r)! * p^(n-r) * q^r
The p is the probability of getting a side A in one toss. Since a counter has only two side, p = 0.5. The q is the probability of not getting side A in one toss, which is also q = 0.5. Now, r is the number of success per n trials. There are 3 tosses so, n=3. The question is getting "at least 1" counter. So, r=1, r=2 and r=3.
Probability for r=1: 3!/1!(3-1)! * (0.5)^(3-1) * (0.5)^1= 0.375
Probability for r=2: 3!/2!(3-2)! * (0.5)^(3-2) * (0.5)^2= 0.375
Probability for r=1: 3!/3!(3-3)! * (0.5)^(3-3) * (0.5)^3= 0.125
Total probability = 0.375 + 0.375 + 0.125 = 0.875
Hello,
We don't need algebra to solve this.
I suppose the ages are 1,2,3,4
Their sum is 1+2+3+4=10.
To obtain 94 we must add 94-10=84 for 4 persons =>84/4=21 to each
Ages are
1+21=22
2+21=23
3+21=24
4+21=25
But if really want an algebric resolution:
Let's x the age of the youngest
x+1,x+2,x+3 the others.
x+(x+1)+(x+2)+(x+3)=94
==>4x+6=94
==>4x=88
==>x=22
Ages are 22,23,24,25.
The oldest friend is 25 years old.
It is about $5.89 for one pair of work gloves. You divide 494.92 by 84 (7 dozen) and that is your unit price.
X is equal to 2 because thirty divide fifteen is two
