Question

A bag contains 2 red, 3 white and 5 black balls, a ball is drawn, its colour is noted and replaced. Minimum number of times, a ball must be drawn so that the probability of getting a red ball for the first time is at least even is?

Answer 1

Hint: To solve this question we should find the combination for various trials until and unless the probability is less than $ \dfrac{1}{2} $ . Be careful here since a number of minimum trials are asked; it can be more than 1 , not like probability.

Complete step-by-step answer:
Total number of balls is 10.
Hence, the probability of drawing a red ball from the bag is $ \dfrac{5}{{10}} = \dfrac{1}{2} $
Now, we need to calculate the probability of drawing at least one red card in 2 trials.
Hence, the probability of drawing red ball from the bag is
 =P(both cards are red)+P(one card is red)
  $
   = \dfrac{1}{5} \times \dfrac{1}{5} + 2 \times \dfrac{1}{5} \times \dfrac{4}{5} \\
   = \dfrac{1}{{25}} + \dfrac{8}{{25}} \\
   = \dfrac{9}{{25}} \;
   $
Now, since the value is less than $ \dfrac{1}{2} $ hence we will check the probability in the third trial.
Hence, the probability of drawing at least one red card in 3 trial from the bag is
=P(all three cards are red)+P(2 cards are red)+P(one card is red)
  $
   = \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{1}{5} + 3 \times \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{4}{5} + 3 \times \dfrac{1}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} \\
   = \dfrac{1}{{125}} + \dfrac{{12}}{{125}} + \dfrac{{48}}{{125}} \\
   = \dfrac{{61}}{{125}} = 0.488 \;
   $
Now, this value is again less than $ \dfrac{1}{2} $ hence we will check the probability for the fourth trial.
Hence, the probability of drawing at least one red card in 4 trial from the bag is
=P(all four cards are red)+P(3 cards are red)+P(2cards are red)+P(1 card is red)
  $
   = \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{1}{5} + 4 \times \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{4}{5} + 6 \times \dfrac{1}{5} \times \dfrac{1}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} + 4 \times \dfrac{1}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} \\
   = \dfrac{1}{{625}} + \dfrac{{16}}{{625}} + \dfrac{{96}}{{625}} + \dfrac{{256}}{{625}} \\
   = \dfrac{{369}}{{625}} = 0.5904 \;
   $
We see that 0.5904 is greater than $ \dfrac{1}{2} $ .
Hence, the minimum number of times a ball must be drawn so that the probability of getting a red ball for the first time is at least even is 4.
So, the correct answer is “4 ”.

Note: We can also use an alternative method to solve this question where we consider the contradiction of the probability.
The probability of not drawing a red ball is $ \dfrac{4}{5} $ .
The probability of not drawing a red ball in n trials is $ {\left( {\dfrac{4}{5}} \right)^n} $ .
Now, the probability that we get at least one red is $ 1 - {\left( {\dfrac{4}{5}} \right)^n} $
This value should be less than $ \dfrac{1}{2} $ .
Hence,
  $
  1 - {\left( {\dfrac{4}{5}} \right)^n} > \dfrac{1}{2} \\
  {\left( {\dfrac{4}{5}} \right)^n} < \dfrac{1}{2} \\
  n\ln \left( {\dfrac{4}{5}} \right) < \ln \left( {\dfrac{1}{2}} \right) \\
  n > \ln \left( {\dfrac{1}{2}} \right)\ln \left( {\dfrac{4}{5}} \right) \simeq 3.1062 \;
   $
Hence, n=4.