Question

A bag contains 6 red shirts,6 green shirts and 8 blue shirts two shirts are drawn randomly.What is the probality that at most one shirt is red​

Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• A bag contains 6 red shirts,6 green shirts and 8 blue shirts two shirts are drawn randomly.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Probability of a red shirt

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Number of red shirts = 6

• Number of green shirts = 6

• Number of blue shirts = 8

Probability of red shirt picked at randam =

• Let R be an event for the probability of red shirt.

We know that,

$\boxed{\pink{\sf P(R) \: = \: \dfrac{Favourable \: outcomes \: for \: event \: R}{Total \: number \: of \: possible \: outcomes}}}$

Here,

• Favourable outcomes for event R = 6

• Total number of possible outcomes = 6 + 6 + 8 = 20

Substituting the values,

• $\sf P(R) \: = \: \dfrac{6}{20}$

• $\sf P(R) \: = \: \dfrac{\cancel{6}}{\cancel{20}}$

• $\sf P(R) \: = \: \dfrac{3}{10}$

• $\sf P(R) \: = \: 0.3$

Therefore,

• $\underline{\boxed{\therefore{\purple{\tt{Probability \: of \: a \: red \: shirt \: = \: \dfrac{3}{10} \: or \: 0.3}}}}}$

$\Large{\bf{\blue{\mathfrak{\dag{\underline{\underline{Extra \: Information:-}}}}}}}$

Probability of an event lies between 0 or 1.

• Let A be any event then probability of that event,

$\sf P(A) \: = \: \dfrac{Favourable \: outcomes \: for \: event \: A}{Total \: number \: of \: possible \: outcomes}$

Answer 2

Answer:

Given :-

• A bag contain 6 red shirts, 6 green shirts and 8 blue shirts.
• Two shirts are drawn randomly.

To Find :-

• What is the probability that at most one shirt is red.

Formula Used :-

$\longmapsto\sf\boxed{\bold{\pink{Probability\: for\: red\: shirt\: =\: \dfrac{The\: total\: favorable\: outcomes\: for\: events\: R}{The\: total\: number\: of\: outcomes}}}}$

Solution :-

Let, the probability that at most one shirt is red be R.

Given :

$\mapsto$ No of red shirts = 6

$\mapsto$ No of green shirts = 6

$\mapsto$ No of blue shirts = 8

First, we have to find the total number of outcomes :

$\implies \sf Red\: shirts\: + Green\: shirts\: + Blue\: shirts$

$\implies \: 6 + 6 + 8$

$\implies \sf 12 + 8$

$\implies \sf\bold{\purple{20}}$

Hence, the total outcomes is 30 .

Given :

• Favourable Outcomes for events R (Red shirts) = 6
• Total number of outcomes = 20

According to the question by using the formula we get,

$\implies \sf Probability\: for\: red\: shirts =\: \dfrac{\cancel{6}}{\cancel{20}}$

$\implies \sf Probability\: for\: red\: shirts =\: \dfrac{\cancel{3}}{\cancel{10}}$

$\implies \sf Probability\: for\: red\: shirts =\: \dfrac{3}{10}$

$\implies\sf\bold{\red{Probability\: for\: red\: shirts =\: 0.3}}$

$\therefore$ The probability that at most one shirt is red is 0.3 .