Two ships a and b are firing cannons at each other. Ship A reaches its target 30% of the time, Ship B 60% of the time. The ships take turns firing at one another. When a ship is hit it sinks. Ship A fires first. What is Ship A's survival probability, rounded to the nearest 0.01? a. 33%.
b. 38%.
c. 42%.
d. 50%.

c. 42%.

Step-by-step explanation:

P(ship A reaches target) : P(A) = 0.30

P(ship B reaches target) : P(B) = 0.60

P( Ship B not reaching target ) : P(B^c) = 1 - p(B) = 1 - 0.60 = 0.40

P( Ship A not reaching target ) : P(A^c) = 1 - p(B) = 1 - 0.30 = 0.70

NOW

define A for A wins

define A^c for A loss

define B for B wins

define B^c for B loss

A ⇒ P(A) = 0.30

A^c B^c A ⇒P(A^c B^c A) = 0.70 × 0.40 × 0.30

A^c B^c A^c B^c A ⇒ P(A^c B^c A^c B^c A) = ( 0.70 × 0.40)² × 0.30

↓↓↓

∴P(A survives) = sum of all probabilities

⇒ 0.30 + ( 0.70 × 0.40 × 0.30 ) + ( 0.70 ×0.40 )² × 0.30 +

⇒ 0.30 [ 1 + ( 0.28) + (0.28)² +]

Now sum of geometric series S∞ = a / 1-β = 1 / 1 - 0.28

∴ 0.30 × 1/1-0.28 = 0.42

now converting to percentage

0.42 × 100 = 42%

Ship A's survival probability is 42%

hence, the required result is

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

we have to compute b-a which means subtracting corresponding elements of a from b.

-

a

step-by-step explanation: