Question

Probability that a hits the target is 4/10 and b is 7/10

Answer 1

Answer:

Step-by-step explanation:

Probability of A hit the target = 4/10;

Probability of B hit the target = 7/10;

The probability hitting the target only once = P(A)[1-P(B)] * P(B)[1- P(A)]

= $\frac{4}{10} (1-\frac{7}{10} ) + \frac{7}{10}(1-\frac{4}{10} )$;

= $\frac{4}{10}*\frac{3}{10} + \frac{7}{10}*\frac{3}{10}$;

The probability hitting the target only once = $\frac{33}{100}$;

Answer 2

Answer:

Step-by-step explanation:

Probability of A hits the target : 4/10

Probi=ability of B hits the target : 7/10

Probability of A not hitting the target : 6/10

Probability of B not hittting the target : 3/10

P ( target is hit once )= P ( A hitting )* P ( B not hitting ) + P ( A not hittin g ) * P ( B hitting)

                              = 4/10 × 3/10 +  6/10 × 7/10

                               = 12 /100 + 42/100

                              = 54 /100

So , the probability of hit at once is 54 /100