Three fair coins are tossed. what is the probability that they are all heads if at least one coin is head,

1/7

Step-by-step explanation:

There are 8 sample points in the sample space. The sample space is

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. The event at least one coin is head is compoused

by seven sample points: {HHH, HHT, HTH, THH, TTH, THT, HTT}, then, to obtain the probability of three heads given that we know that at least one coin is head, we should regard the "new sample space" reduced, i.e., we should consider the set with seven sample points as the new sample space. In the new sample space, just one point corresponds to three heads. Therefore the probability we are seeking is 1 out of 7 or 1/7.

answer: 15

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .