Translate left-handedness counts into binomial probabilities

Let X count the left-handed people among the 12 selected. Use $X\sim B(12,0.15)$. The source lists all four conditions: 12 repeated trials; left-handed/not-left-handed categories; constant success probability 0.15; and independence of each person’s handedness in the sampling model. For sampling without replacement from a large finite population, this independence is an approximation justified when the sample is negligible relative to the population; random selection alone does not make arbitrary samples independent.

$P(X=2)=\binom{12}2(0.15)^2(0.85)^{10}=0.292$ to 3 significant figures.

1. Press MODE and set STAT WIZARDS: ON to activate the statistics input wizard. 2. Press 2nd, VARS; select A:binompdf( and ENTER. The source expands the name as binomial probability distribution function; here it computes one probability mass. 3. Enter trials:12, p:0.15, x value:2, highlight Paste, then ENTER. 4. Press ENTER again. The home screen shows binompdf(12,0.15,2) with output 0.2923584904, hence 0.292 to 3 significant figures.

$P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$. Retain all four source terms: $\binom{12}0(0.15)^0(0.85)^{12}+\binom{12}1(0.15)^1(0.85)^{11}+\binom{12}2(0.15)^2(0.85)^{10}+\binom{12}3(0.15)^3(0.85)^9=0.908$ to 3 significant figures.

1. Press 2nd, VARS; select B:binomcdf( and ENTER. The command name means binomial cumulative distribution function and computes $P(X\le x)$. 2. Enter trials:12, p:0.15, x value:3, choose Paste, then ENTER. 3. Press ENTER again. The source home screen is binomcdf(12,.15,3) with output 0.9077936688, hence 0.908 to 3 significant figures.

$P(X>4)=1-P(X\le4)=0.0239$ to 3 significant figures. Subtract the cumulative probability at 4, not at 3.

Twelve minus X is the number who are not left-handed. Thus $12-X\ge7$ means $X\le5$. The required probability is $P(X\le5)=0.995$ to 3 significant figures.

The source explicitly uses strict endpoints: $P(3<X<9)=P(X\le8)-P(X\le3)=0.0922$ to 3 significant figures. Its later remark says “between” excludes the endpoints; retain this convention for this question, but use an explicitly stated inequality when another problem specifies inclusive endpoints.