Question

A bag contains 6 Red balls and 4 yellow balls. Two balls are picked at random what is the probability that at least one is red?

Answer 1

Answer:

1/3

Step-by-step explanation:

Initially, there are 6 red balls and 4 yellow balls.

P(R1) = Probability of getting a red ball on the first draw = (6 / 10).

If the first draw yields a red ball; there are 5 red balls and 4 yellow balls left.

P(R2) = Probability of getting a red ball on the first draw = (5 / 9).

Therefore, probability of getting two red balls from the first two draws

= P(R1)*P(R2) = (6 / 10)*(5 / 9) = (1 / 3) = 0.3333.

Answer 2

The probability that at least one ball is red = 13/15

Given :

• A bag contains 6 Red balls and 4 yellow balls.

• Two balls are picked at random

To find :

The probability that at least one is red

Solution :

Step 1 of 3 :

Find total number of possible outcomes

The bag contains 6 Red balls and 4 yellow balls.

Total number of balls = 6 + 4 = 10

Now two balls are selected at random

So total number of possible outcomes

$\displaystyle \sf = {}^{10} C_2$

= 45

Step 2 of 3 :

Find the probability that no ball is is red

Let A be the event that no ball is is red

Therefore two balls which are picked at random must be both yellows

Total number of possible outcomes for the event A

$\displaystyle \sf = {}^{4} C_2$

= 6

∴ The probability that no ball is is red

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{6}{45} }$

$\displaystyle \sf{ = \frac{2}{15} }$

Step 3 of 3 :

Find the probability that at least one is red

Hence the required probability that at least one is red

$\displaystyle \sf = P( \bar{A})$

$\displaystyle \sf = 1 - P(A)$

$\displaystyle \sf{ =1- \frac{2}{15} }$

$\displaystyle \sf{ = \frac{13}{15} }$

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