If three unbiased coins are tossed, then what is the probability of getting one or more heads is

The odds would be 7/8 that at least one would be a heads.

To find your answer, see each coin flip as unrelated to the last one.

For the first coin flip, you can either get heads or tails. So, there's only 2 possible outcomes.

Flip 1 (Outcome 1): H or T

Flip 1 (Outcome 2): H or T

But when it comes to the second flip, now each outcome has two possible "sub-outcomes" of their own. So there's now four total outcomes.

Flip 2 (Outcome 1) : H, H

Flip 2 (Outcome 2) : H, T

Flip 2 (Outcome 3) : T, H

Flip 2 (Outcome 4) : T, T

On the third flip, there would now be 8 outcomes. As each "sub-outcome" has own of their own.

Flip 3 (Outcome 1) : H, H, H

Flip 3 (Outcome 2) : H, H, T

Flip 3 (Outcome 3) : H, T, H

Flip 3 (Outcome 4 : H, T, T

Flip 3 (Outcome 5) : T, H, H

Flip 3 (Outcome 6) : T, H, T

Flip 3 (Outcome 7) : T, T, H

Flip 3 (Outcome 8): T, T, T

So all but one of the eight outcomes have at least one heads drawn. Due to this, the odds of an heads happening would be 7/8 within all of the possible outcomes.

answer: group b.

step-by-step explanation:

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.