Mathematics, 21.06.2020 02:57 myaj580

b. Find the probability that two or fewer heads are observed in three tosses. (Round your answer to three decimal places.) c. Find the probability that at least one head is observed in three tosses. (Round your answer to three decimal places.) d. Find the expected value of X. (Round your answer to one decimal place.) e. Find the standard deviation of X. (Round your answer to three decimal places.)

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Answer from: scavalieri2421

(a) Probability distribution is prepared below.

(b) The probability that two or fewer heads are observed in three tosses is 0.875.

(d) The expected value of X is 1.5.

(e) The standard deviation of X is 2.121.

Step-by-step explanation:

The complete question is: A fair coin is tossed three times. Let X be the number of heads observed in three tosses of this fair coin.

(a) Find the probability distribution of X.

(b) Find the probability that two or fewer heads are observed in three tosses. (Round your answer to three decimal places.)

(c) Find the probability that at least one head is observed in three tosses. (Round your answer to three decimal places.)

(d) Find the expected value of X. (Round your answer to one decimal place.)

(e) Find the standard deviation of X. (Round your answer to three decimal places.)

Now, firstly the sample space obtained in three tosses of a fair coin is given as;

Sample Space (S) = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}

(a) The Probability distribution of X is given below;

Number of Heads (X) P(X) $X \times P(X)$ $X^{2} \times P(X)$

0 $\frac{1}{8}$ = 0.125 0 0

1 $\frac{3}{8}$ = 0.375 0.375 0.375

2 $\frac{3}{8}$ = 0.375 0.75 3

3 $\frac{1}{8}$ = 0.125 0.375 3.375

Total 1.5 6.75

(b) The probability that two or fewer heads are observed in three tosses is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 0.125 + 0.375 + 0.375

= 0.875

P(X $\geq$ 1) = 1 - P(X = 0)

= 1 - 0.125

= 0.875

(d) The expected value of X = E(X) = $\sum (X \times P(X))$

= 1.5

(e) The Variance of X = V(X) = $E(X^{2} ) - ( E(X))^{2}$

= $\sum (X^{2} \times P(X))- (\sum (X \times P(X)))^{2}$

= $6.75 - 1.5^{2}$ = 4.5

Now, Standard deviation of X = $\sqrt{V(X)}$

= $\sqrt{4.5}$ = 2.121.

Answer from: Quest

It is blindness and randomization

Answer from: Quest

n=1.75+1.75

step-by-step explanation:

3.5/2=1.75

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