A decret random variable X is such that X~B(8, 1/4). Find a) P(X=2) b) P(X greater Dan or equal to 1)

Computer Science

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Answers

2 parts
a

510316384\frac{5103}{16384}

b

5897565536\frac{58975}{65536}

Step-by-step solution

7 steps

The random variable XX follows a binomial distribution, X∼B(n,p)X \sim B(n, p), where n=8n=8 (number of trials) and p=1/4p=1/4 (probability of success). The probability mass function is given by:

P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-k}

Here, n=8n=8, p=1/4p=1/4, and 1−p=1−1/4=3/41-p = 1 - 1/4 = 3/4.

Part a· 4 steps

Find P(X=2)P(X=2)

  1. Identify the values for nn, kk, and pp.

    For P(X=2)P(X=2), we have n=8n=8, k=2k=2, p=1/4p=1/4, and 1−p=3/41-p=3/4.

  2. Substitute the values into the binomial probability formula.

    P(X=2)=(82)(14)2(34)8−2P(X=2) = \binom{8}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^{8-2} P(X=2)=(82)(14)2(34)6P(X=2) = \binom{8}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^6

  3. Calculate the binomial coefficient (82)\binom{8}{2}.

    (82)=8!2!(8−2)!=8!2!6!=8×72×1=28\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28

  4. Calculate the powers and multiply.

    P(X=2)=28×(116)×(3646)P(X=2) = 28 \times \left(\frac{1}{16}\right) \times \left(\frac{3^6}{4^6}\right) P(X=2)=28×116×7294096P(X=2) = 28 \times \frac{1}{16} \times \frac{729}{4096} P(X=2)=28×72916×4096=7×7294×4096=510316384P(X=2) = \frac{28 \times 729}{16 \times 4096} = \frac{7 \times 729}{4 \times 4096} = \frac{5103}{16384} P(X=2)≈0.31146P(X=2) \approx 0.31146

    The probability is 510316384\boxed{\frac{5103}{16384}}.

Part b

5897565536\frac{58975}{65536}

Find P(X≥1)P(X \ge 1)

  1. Use the complement rule.

    It is easier to calculate P(X≥1)P(X \ge 1) as 1−P(X<1)1 - P(X < 1). Since XX is a discrete random variable, P(X<1)P(X < 1) is equivalent to P(X=0)P(X=0).

    P(X≥1)=1−P(X=0)P(X \ge 1) = 1 - P(X=0)

  2. Calculate P(X=0)P(X=0).

    For P(X=0)P(X=0), we have n=8n=8, k=0k=0, p=1/4p=1/4, and 1−p=3/41-p=3/4.

    P(X=0)=(80)(14)0(34)8−0P(X=0) = \binom{8}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^{8-0} P(X=0)=1×1×(34)8P(X=0) = 1 \times 1 \times \left(\frac{3}{4}\right)^8 P(X=0)=3848=656165536P(X=0) = \frac{3^8}{4^8} = \frac{6561}{65536}

  3. Subtract P(X=0)P(X=0) from 1.

    P(X≥1)=1−656165536P(X \ge 1) = 1 - \frac{6561}{65536} P(X≥1)=65536−656165536=5897565536P(X \ge 1) = \frac{65536 - 6561}{65536} = \frac{58975}{65536} P(X≥1)≈0.90003P(X \ge 1) \approx 0.90003

    The probability is 5897565536\boxed{\frac{58975}{65536}}.

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