Question

A box contains $100$red cards, $200$yellow cards and $50$blue cards if a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card.

Answer 1

Probability:- In our daily life, such a situation may happen everyday when we guess about the happening of an event.
The likelihood or chance of the happening of an event is termed as probability.
If an event can happen in ‘$m$’ ways and fails to happen in ‘$n$’ ways and each of these $(m + n)$ ways is equally likely to occur, then the probability of the happening of the event is
$p = \dfrac{m}{{m + n}}$ , where probability is denoted by ‘$p$’
and the probability of not happening the event is
$q = 1 - \dfrac{m}{{m + n}} = \dfrac{n}{{m + n}}$
Or
$q = 1 - p$
Also, we have
$p + q = 1$

Complete step by step solution:
We have,
Red cards$= 100$
Yellow cards$= 200$
Blue cards$= 50$
Total number of cards $= 100 + 200 + 50$
$= 350$

(i) A blue card:
Number of blue cards $= 50$
P (drawing a blue card) $= \dfrac{{50}}{{350}}$
$= \dfrac{1}{7}$

(ii) Not a yellow card:
Number of yellow cards which are not yellow $= 350 - 200$
$= 150$
$\therefore$ P (drawing a card that is not yellow) $= \dfrac{{150}}{{350}}$
$= \dfrac{3}{7}$

(iii) Neither yellow nor a blue card:
Number of yellow or blue card $= 200 + 50$
$= 250$
P (drawing a yellow or blue card) $= \dfrac{{250}}{{350}} = \dfrac{5}{7}$
q (neither yellow nor blue card) $= 1 - P$
$= 1 - \dfrac{5}{7}$
$= \dfrac{{7 - 5}}{7}$
$= \dfrac{2}{7}$
Hence, the probability of getting a blue card, not a yellow card and neither yellow nor blue card are $\dfrac{1}{7}$,$\dfrac{3}{7}$and $\dfrac{2}{7}$ respectively.

Note: General formula of Probability
P [E] $= \dfrac{{No.\,of\,favourable\,outcomes}}{{Total\,no.\,of\,Possible\,outcomes}}$