The two shooters shoot at the target once. the probability of both hitting is 0.54, the probability of both being missed

The two shooters shoot at the target once. the probability of both hitting is 0.54, the probability of both being missed is 0.04, what is the probability of hitting the target by each shooter with one shot?

1. Suppose that the probabilities of events A1 and A2, consisting in the fact that each of the shooters will hit the target, are equal, respectively:

p1 = P (A1);
p2 = P (A2).
2. Then the probabilities of opposite events A1 ‘and A2’ that they will miss are equal, respectively:

q1 = P (A1 ‘) = 1 – P (A1) = 1 – p1;
q2 = P (A2 ‘) = 1 – P (A2) = 1 – p2.
3. The probability of event X that both of them hit the target:

P (X) = P (A1) * P (A2) = p1p2.
And the probability of event Y that they both miss is:

P (Y) = P (A1 ‘) * P (A2’) = q1q2.
4. Based on these formulas, we compose the equations:

{p1p2 = 0.54;
{q1q2 = 0.04;
{p1p2 = 0.54;
{(1 – p1) (1 – p2) = 0.04;
{p1p2 = 0.54;
{1 + p1p2 – p1 – p2 = 0.04;
{p1p2 = 0.54;
{1 + 0.54 – p1 – p2 = 0.04;
{p1p2 = 0.54;
{p1 + p2 = 1.5.
5. By Vieta’s theorem, p1 and p2 are the roots of the quadratic equation:

t ^ 2 – 1.5t + 0.54 = 0;
D = 1.5 ^ 2 – 4 * 0.54 = 2.25 – 2.16 = 0.09;
t = (1.5 ± √0.09) / 2 = (1.5 ± 0.03) / 2;
t1 = (1.5 – 0.03) / 2 = 1.47 / 2 = 0.735;
t2 = (1.5 + 0.03) / 2 = 1.53 / 2 = 0.765.
Answer: 0.735; 0.765.