问题如下图：
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解释：
老师你好，请问下怎么判断这个事件是遵从Binomial Distribution的呢，就是可能一下子想不到要用到这个公式，谢谢
NO.PZ2017092702000082 问题如下 A portfolio manager annually outperforms her benchmark 60% of the time. Assuming inpennt annutrials, whis the probability thshe will outperform her benchmark four or more times over the next five years? A.0.26 B.0.34 C.0.48 B is correct.To calculate the probability of 4 years of outperformance, use the formulp(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx)px(1−p)n−x Using this formula to calculate the probability in 4 of 5 years, n = 5, x = 4 anp = 0.60. Therefore, p(4)=5!(5−4)!4!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p{(4)}=\frac{5!}{{(5-4)}!4!}0.6^4{(1-0.6)}^{5-4}={(120/24)}{(0.1296)}{(0.40)}=0.2592p(4)=(5−4)!4!5!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p(5)=5!(5−4)!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778p{(5)}=\frac{5!}{{(5-4)}!5!}0.6^5{(1-0.6)}^{5-5}={(120/120)}{(0.0778)}{(1)}=0.0778p(5)=(5−4)!5!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778The probability of outperforming 4 or more times is p(4) + p(5) = 0.2592 + 0.0778 = 0.3370 计算outperform四次的公式如下5C4*0.6^4为什么不是后面乘以5C1（1-0.6），而是只乘以1（1-0.6）
NO.PZ2017092702000082问题如下 A portfolio manager annually outperforms her benchmark 60% of the time. Assuming inpennt annutrials, whis the probability thshe will outperform her benchmark four or more times over the next five years?A.0.26B.0.34C.0.48B is correct.To calculate the probability of 4 years of outperformance, use the formulp(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx)px(1−p)n−x Using this formula to calculate the probability in 4 of 5 years, n = 5, x = 4 anp = 0.60. Therefore, p(4)=5!(5−4)!4!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p{(4)}=\frac{5!}{{(5-4)}!4!}0.6^4{(1-0.6)}^{5-4}={(120/24)}{(0.1296)}{(0.40)}=0.2592p(4)=(5−4)!4!5!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p(5)=5!(5−4)!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778p{(5)}=\frac{5!}{{(5-4)}!5!}0.6^5{(1-0.6)}^{5-5}={(120/120)}{(0.0778)}{(1)}=0.0778p(5)=(5−4)!5!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778The probability of outperforming 4 or more times is p(4) + p(5) = 0.2592 + 0.0778 = 0.3370这道题计算没有错误么 第一步得出来是0.5184呀C54是10*0.6^4*0.4=0.5184
NO.PZ2017092702000082 问题如下 A portfolio manager annually outperforms her benchmark 60% of the time. Assuming inpennt annutrials, whis the probability thshe will outperform her benchmark four or more times over the next five years? A.0.26 B.0.34 C.0.48 B is correct.To calculate the probability of 4 years of outperformance, use the formulp(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx)px(1−p)n−x Using this formula to calculate the probability in 4 of 5 years, n = 5, x = 4 anp = 0.60. Therefore, p(4)=5!(5−4)!4!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p{(4)}=\frac{5!}{{(5-4)}!4!}0.6^4{(1-0.6)}^{5-4}={(120/24)}{(0.1296)}{(0.40)}=0.2592p(4)=(5−4)!4!5!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p(5)=5!(5−4)!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778p{(5)}=\frac{5!}{{(5-4)}!5!}0.6^5{(1-0.6)}^{5-5}={(120/120)}{(0.0778)}{(1)}=0.0778p(5)=(5−4)!5!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778The probability of outperforming 4 or more times is p(4) + p(5) = 0.2592 + 0.0778 = 0.3370 求问公式里的( n x )在答案里是怎么体现出来的啊？
老师您好，看不太懂答案，可以一下具体解题步骤和计算器步骤吗？
不懂公式，尤其是感叹号那块儿，求讲解，以及如何一步一步的按计算器，以及困扰了两天了。谢谢！
