If a baseball player has a batting average of 0.275, what is the likelihood that the player will get the following number of hits in the next four times at bat?(A) Exactly 2 hits&nbsp;(B) At least 2 hits&nbsp;(A) P(exactly 2 hits)&nbsp;≈.029&nbsp;(Round to three decimal places as needed.)

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If a baseball player has a batting average of 0.275, what is the likelihood that the player will get the following number of hits in the next four times at bat?(A) Exactly 2 hits&amp;nbsp;(B) At least 2 hits&amp;nbsp;(A) P(exactly 2 hits)&amp;nbsp;≈.029&amp;nbsp;(Round to three decimal places as needed.)

nghodlokl · Accepted Answer

Step 1Introduction:Denote X as the random variable representing the number of hits made by the player in the next 4 times at bat.Step 2A. Consider that the player hits or misses each of the next 4 times at bat, independently of each other. Further, consider it to be a success if the player makes a hit.As the batting average is 0.275, the probability of “success” is&amp;nbsp;p=0.275.Here, X can be considered as the number of successes observed among the 4 times at bat. Thus, it may be assumed that, X has a binomial distribution with parameters,&amp;nbsp;n=4,p=0.275.Since p denotes the probability of success, the probability of failure is,&amp;nbsp;q=1-p=0.725.If&amp;nbsp;X~Binomialn,p, then the probability mass function of X is:f(x)={(nx)pxqn−x;x=0,1,...,n;&amp;amp;0&amp;nbsp;Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.The probability mass function of X here is:f(x)={(4x)(0.275)x(0.725)4−x;x=0,1,2,3,4.0;otherwiseThe probability that the player will get exactly 2 hits, that is,&amp;nbsp;PX=2&amp;nbsp;is calculated below:PX=2=f2=(42)(0.275)2(0.725)4−2=6·0.075625·0.525625≈0.239.Thus, the probability that the player will get exactly 2 hits is 0.239.Step 3B. The probability that the player will get at least 2 hits, that is,&amp;nbsp;PX≥2&amp;nbsp;is calculated below:PX≥2=f2+f3+f4=(42)(0.275)2(0.725)4−2+(43)(0.275)3(0.725)4−3+(44)(0.275)4(0.725)4−4(≈(6)⋅(0.075626)⋅(0.525625)+(4)⋅(0.020797)⋅(0.725)+(1)⋅(0.2754)⋅(1)≈0.238502+0.060311+0.005719≈0.305.Thus, the probability that the player will get at least 2 hits is 0.305.

Jim Hunt · Accepted Answer

Step 1  Here we are given X∼Binomial(μ, 0.215)  so we are to find  a) P(x=2)=4cx(0.275)2(0.725)2=0.2385  b) P(x≥2)=1−∑x=024cx(0.275)x(0.725)4−x  =0.5808  as P(x=x)=ncx+x(1−p)n−x

nick1337 · Accepted Answer

Step 1 Let X the random variable of interest, on this case we now that: X∼Binom(n=4, p=0.375) The probability mass function for the Binomial distribution is given as: P(X)=(nCx)(p)x(1−p)n−x Where (nCx) means combinatory and it&#039;s given by this formula: nCx=n!(n−x)!x! Part a P(X=2)=(4C2)(0.375)2(1−0.375)4−2=0.330 Part b P(X≥2)=1−P(X&amp;lt;2)=1−[P(X=0)+P(X=1)] P(X=0)=(4C0)(0.375)0(1−0.375)4−0=0.153 P(X=1)=(4C1)(0.375)1(1−0.375)4−1=0.366 And replacing we got: P(X≥2)=1−P(X&amp;lt;2)=1−[0.153+0.3660=0.481

If a baseball player has a batting average of 0.275, what is

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