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Two red balls and two white balls are placed in a bag ,balls are drawn one by one at random without replacement random variable x is the number of white balls drawn before the first red ball is shown

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Answers

2 parts
b

E(X)=23E(X) = \frac{2}{3}

Var(X)=59Var(X) = \frac{5}{9}

c

P(Xˉ>0.75)≈0.3618P(\bar{X} > 0.75) \approx 0.3618

Step-by-step solution

8 steps

Here's a breakdown of the problem:

The bag contains 2 red balls (R) and 2 white balls (W), for a total of 4 balls. Balls are drawn one by one without replacement.
The random variable XX is the number of white balls drawn before the first red ball is shown.
Possible values for XX are 0, 1, or 2.

Part a· 2 steps

Show how the result P(X=1)=3/10P(X=1) = 3/10 is obtained, and find the rest of the probability distribution of XX.

  1. Determine the possible sequences of draws and their probabilities.

    • X=0X=0: The first ball drawn is Red. P(X=0)=P(1st is R)=24=12P(X=0) = P(\text{1st is R}) = \frac{2}{4} = \frac{1}{2}

    • X=1X=1: The first ball drawn is White, and the second ball drawn is Red. P(X=1)=P(1st is W)×P(2nd is R | 1st is W)P(X=1) = P(\text{1st is W}) \times P(\text{2nd is R | 1st is W}) P(1st is W)=24P(\text{1st is W}) = \frac{2}{4} After drawing one white ball, there are 3 balls left (1W, 2R). P(2nd is R | 1st is W)=23P(\text{2nd is R | 1st is W}) = \frac{2}{3} P(X=1)=24×23=412=13P(X=1) = \frac{2}{4} \times \frac{2}{3} = \frac{4}{12} = \frac{1}{3}

    • X=2X=2: The first ball is White, the second ball is White, and the third ball is Red. P(X=2)=P(1st is W)×P(2nd is W | 1st is W)×P(3rd is R | 1st W, 2nd W)P(X=2) = P(\text{1st is W}) \times P(\text{2nd is W | 1st is W}) \times P(\text{3rd is R | 1st W, 2nd W}) P(1st is W)=24P(\text{1st is W}) = \frac{2}{4} After drawing one white ball, there are 3 balls left (1W, 2R). P(2nd is W | 1st is W)=13P(\text{2nd is W | 1st is W}) = \frac{1}{3} After drawing two white balls, there are 2 balls left (0W, 2R). P(3rd is R | 1st W, 2nd W)=22=1P(\text{3rd is R | 1st W, 2nd W}) = \frac{2}{2} = 1 P(X=2)=24×13×1=212=16P(X=2) = \frac{2}{4} \times \frac{1}{3} \times 1 = \frac{2}{12} = \frac{1}{6}

  2. Summarize the probability distribution.

    X012P(X=x)121316\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X=x) & \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \hline \end{array}

    Check sum: 12+13+16=36+26+16=66=1\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1.

    Note on the problem statement: The problem asks to "show that probability P(x) = 1 = 30". This appears to be a typo and likely meant "show that P(X=1)=3/10P(X=1) = 3/10". However, based on the problem description (2 red, 2 white balls), our calculation shows P(X=1)=1/3P(X=1) = 1/3. The value 3/103/10 is inconsistent with the given setup. We proceed with our calculated value of 1/31/3.

Part b

E(X)=23E(X) = \frac{2}{3}

Find E(X)E(X) and show that Var(X)=5/9Var(X) = 5/9.

  1. Calculate the expected value E(X)E(X).

    E(X)=∑xP(X=x)E(X) = \sum x P(X=x) E(X)=0(12)+1(13)+2(16)E(X) = 0 \left(\frac{1}{2}\right) + 1 \left(\frac{1}{3}\right) + 2 \left(\frac{1}{6}\right) E(X)=0+13+26E(X) = 0 + \frac{1}{3} + \frac{2}{6} E(X)=13+13E(X) = \frac{1}{3} + \frac{1}{3} E(X)=23\boxed{E(X) = \frac{2}{3}}

  2. Calculate E(X2)E(X^2).

    E(X2)=∑x2P(X=x)E(X^2) = \sum x^2 P(X=x) E(X2)=02(12)+12(13)+22(16)E(X^2) = 0^2 \left(\frac{1}{2}\right) + 1^2 \left(\frac{1}{3}\right) + 2^2 \left(\frac{1}{6}\right) E(X2)=0+13+4(16)E(X^2) = 0 + \frac{1}{3} + 4 \left(\frac{1}{6}\right) E(X2)=13+46E(X^2) = \frac{1}{3} + \frac{4}{6} E(X2)=13+23E(X^2) = \frac{1}{3} + \frac{2}{3} E(X2)=1E(X^2) = 1

  3. Calculate the variance Var(X)Var(X).

    Var(X)=E(X2)−(E(X))2Var(X) = E(X^2) - (E(X))^2 Var(X)=1−(23)2Var(X) = 1 - \left(\frac{2}{3}\right)^2 Var(X)=1−49Var(X) = 1 - \frac{4}{9} Var(X)=99−49Var(X) = \frac{9}{9} - \frac{4}{9} Var(X)=59\boxed{Var(X) = \frac{5}{9}}

    This matches the value given in the problem statement.

Part c

P(Xˉ>0.75)≈0.3618P(\bar{X} > 0.75) \approx 0.3618

A sample of 10 independent observations of XX is denoted by Xˉ\bar{X}. Use the Central Limit Theorem to estimate P(Xˉ>0.75)P(\bar{X} > 0.75).

Given:

  • Sample size, n=10n = 10
  • Population mean, μ=E(X)=23\mu = E(X) = \frac{2}{3}
  • Population variance, σ2=Var(X)=59\sigma^2 = Var(X) = \frac{5}{9}
  • Population standard deviation, σ=59=53\sigma = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}

  1. Calculate the mean and standard deviation of the sample mean Xˉ\bar{X}.

    According to the Central Limit Theorem:

    μXˉ=μ=23\mu_{\bar{X}} = \mu = \frac{2}{3} σXˉ=σn=5310\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{\frac{\sqrt{5}}{3}}{\sqrt{10}} σXˉ=5310=5352=132\sigma_{\bar{X}} = \frac{\sqrt{5}}{3\sqrt{10}} = \frac{\sqrt{5}}{3\sqrt{5}\sqrt{2}} = \frac{1}{3\sqrt{2}}

    To rationalize the denominator:

    σXˉ=132×22=26\sigma_{\bar{X}} = \frac{1}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{6} σXˉ≈0.2357\sigma_{\bar{X}} \approx 0.2357

  2. Standardize Xˉ\bar{X} to a z-score.

    We want to find P(Xˉ>0.75)P(\bar{X} > 0.75).

    Z=Xˉ−μXˉσXˉZ = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} Z=0.75−2326Z = \frac{0.75 - \frac{2}{3}}{\frac{\sqrt{2}}{6}}

    Convert 0.75 to a fraction: 0.75=340.75 = \frac{3}{4}.

    Z=34−2326=912−81226Z = \frac{\frac{3}{4} - \frac{2}{3}}{\frac{\sqrt{2}}{6}} = \frac{\frac{9}{12} - \frac{8}{12}}{\frac{\sqrt{2}}{6}} Z=11226=112×62Z = \frac{\frac{1}{12}}{\frac{\sqrt{2}}{6}} = \frac{1}{12} \times \frac{6}{\sqrt{2}} Z=6122=122Z = \frac{6}{12\sqrt{2}} = \frac{1}{2\sqrt{2}}

    Rationalize the denominator:

    Z=24≈0.3536Z = \frac{\sqrt{2}}{4} \approx 0.3536

  3. Find the probability using the standard normal distribution.

    We need to find P(Z>0.3536)P(Z > 0.3536).
    Using a standard normal distribution table or calculator:

    P(Z>0.3536)=1−P(Z≤0.3536)P(Z > 0.3536) = 1 - P(Z \le 0.3536) P(Z≤0.3536)≈0.6382P(Z \le 0.3536) \approx 0.6382 P(Z>0.3536)≈1−0.6382P(Z > 0.3536) \approx 1 - 0.6382 P(Xˉ>0.75)≈0.3618\boxed{P(\bar{X} > 0.75) \approx 0.3618}
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