Tanya prepared 4 different letters to be sent to 4 different addresses. for each letter, she prepared an envelope with its correct address. if the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

Answer:

The probability that only 1 letter will be put into the envelope with its correct address is

Step-by-step explanation:

Given:

Number of Letters=4

Number of addresses= 4

To Find:

The probability that only 1 letter will be put into the envelope with its correct address=?

Solution:

Let us assume first letter goes in correct envelope and others go in wrong envelopes, then

=> Probability putting the first letter in correct envelope =

=>  Probability putting the second letter in correct envelope =

=>  Probability putting the third letter in correct envelope=

=> Probability putting the fourth letter in correct envelope = 1;  

( only 1 wrong addressed envelope is left);

This event can occur with other 3 envelopes too.

Hence total prob. =

=>

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.