Question

A box contains 3 defective mangoes and 21 good mangoes. one mangoes drawn from the box at random. find the probability of getting
i) a deffective mangoe
ii)a good mangoe

Answer 1

probability of getting a defective mango ==


defective mangoes÷total mangoes
==
3/21+3=3/24=1/8


answer is 1/8








probability of good mangoes==

good mangoes÷total mangoes
==

21/24=7/8



probability of good mangoes = 7/8

Answer 2

i) $\text{P(d)}=\frac{1}{8}$

ii) $\text{P(g)}=\frac{7}{8}$

Step-by-step explanation:

Given : A box contains 3 defective mangoes and 21 good mangoes. one mangoes drawn from the box at random.

To find : The probability of getting

i) a defective mangoes

ii) a good mangoes

Solution :

A box contains 3 defective mangoes and 21 good mangoes.

Total number of mangoes = 3+21=24

i) Favorable outcome of getting  a defective mangoes  = 3

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total outcome}}$

$\text{P(d)}=\frac{3}{24}$

$\text{P(d)}=\frac{1}{8}$

ii) Favorable outcome of getting  a good mangoes  = 21

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total outcome}}$

$\text{P(g)}=\frac{21}{24}$

$\text{P(g)}=\frac{7}{8}$

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