Question

A husband and wife appear in an interview for two vacancies for the same post. Probability of husband getting selected is 1/7 and that of wife getting selected is 1/5 . What is probabliitythat
1. Both os them will be selected.?
2. Only one of them will be selected.?
3. None of them will be selected?

Answer 1

Here's ur answer my frd:

a)1/7*1/5 for both
b)1-(1-1/7-1/5)-(1/7*1/5) only one not both and not none 
c)1-1/7-1/5 


Hope This Helps :)

Answer 2

Answer:  1. $\dfrac{1}{35}$

2.  $\dfrac{2}{7}$

3. $\dfrac{24}{35}$

Step-by-step explanation:

Given : A husband and wife appear in an interview for two vacancies for the same post.

Let H denotes the event of selecting husband and W denotes the event of selecting wife.

Probability of husband getting selected is $P(H)=\dfrac{1}{7}$ .

∴ Probability of husband not getting selected is $P(not\ H)=1-\dfrac{1}{7}=\dfrac{6}{7}$ .

Probability of wife getting selected is $P(W)=\dfrac{1}{5}$ .

∴ Probability of wife not getting selected is $P(not\ W)=1-\dfrac{1}{5}=\dfrac{4}{5}$ .

1. Probability that both of them will be selected :_

$P(H)\times P(W)=\dfrac{1}{7}\times\dfrac{1}{5}=\dfrac{1}{35}$

2. Probability that only one of them will be selected :-

$P(H)\times P(not\ W)+P(W)\times P(not\ H)\\\\=\dfrac{1}{7}\times\dfrac{4}{5}+\dfrac{1}{5}\times\dfrac{6}{7}\\\\=\dfrac{10}{35}=\dfrac{2}{7}$

3.  Probability that none of them will be selected :-

$P(Not\ H)\times P(not\ W)\\\\=\dfrac{6}{7}\times\dfrac{4}{5}=\dfrac{24}{35}$