A code consists of two digit numbers chosen from 00-99 (i. E 00, 01, 02…..99) followed by two different letters of the alphabet. What is the probability that the code is 43MZ?

1/67600

Step-by-step explanation:

probability that the code is 43MZ:

Probability = (Total Required outcome / total possible outcomes)

Possible outcomes (digit) = (00, ..., 99)

2.) possible outcomes (alphabet) = (A, B, .., Z)

P(43MZ) = P(43) × P(M) × P(Z)

P(43) = 1/100

P(M) = 1/26

P(Z) = 1/26

THEREFORE ;

P(43MZ) = (1/100) × (1/26) × (1/26) = 1/67600

did you get the answer yet?

step-by-step explanation:

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4