Question

A survey shows that after production 5% of a packet of chips are underweight. If a sample of 20 packets were selected randomly, find the probability that at most 2 packets are underweight.

Answer 1

solution

From the binomial distribution;

$$P(X=r)=\ _nC_r *p^r(1-p)^{n-r}$$

Where p is the probability of success, n is the number of trials and r, the number of successes in the trials

Let the Probability of getting underweight chips be be P:

$$p =5\%$$

The probability that atmost 2 are underweight is the probability of getting 0, 1 or 2 underweight chips.

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

P( X = 0):

$$P(X=0)=\ _{20}C_0 *0.05^0(1-0.05)^{20-0}$$ $$=0.3585$$

P(X = 1):

$$P(X=1)=\ _{20}C_1*0.05^1(1-0.05)^{20-1}$$ $$=0.3775$$

P(X = 2):

$$P(X=2)=\ _{20}C_2 *0.05^2(1-0.05)^{20-2}$$ $$=0.1887$$

Therefore P( X $\leq$ 2):

$$0.3585+0.3775+0.1887 =0.9245$$

answer: the probability of getting at most 2 underweight chips is 0.9245