Question

Pat is required to sell candy bars to raise money for the 6th grade field trip. There is a 40%

Answer 1

Probabilities are used to determine the chances of events.

The given parameters:

$n = 5$ ---- the number of candy bars

$p = 40\%$ ---- the probability of selling a candy bar

(a) Name distributions

The distribution of X is represented as:

$X \sim(r,p)$

Where:

$r = 5$

$p= \frac{5\times 40\%}{10}$

$p= 0.2$

So, the name distribution of X is $X \sim(r= 5,p = 0.2)$

(b) The probability that the last candy is sold at the 11th house

This means that:

$n = 10$ --- the number of previous houses

$r = 4$ --- the previous number of candies

$p = 0.4$ --- the given probability of selling a candy

The probability is calculated using:

$P(x = n+1) = ^{n}C_r \times p^{r +1} \times (1 - p)^{n-r}$

This gives

$P(x = 10+1) = ^{10}C_4 \times 0.4^{4 +1} \times (1 - 0.4)^{10-4}$

$P(x = 11) = ^{10}C_4 \times 0.4^{5} \times (0.6)^6$

$P(x = 11) = 210 \times 0.4^5 \times 0.6^6$

$P(x = 11) = 0.1003290624$

Approximate

$P(x = 11) = 0.1003$

Hence, the probability that the last candy is sold at the 11th house is 0.1003

(b) The probability he sells the candies on or before the 8th house

The probability is calculated using:

$P(x \le 8) = P(5 \le x \le 8)$

This gives

$P(x \le 8) = ^{10}C_5 \times 0.4^{6} \times (0.6)^5 + ^{10}C_6 \times 0.4^{7} \times (0.6)^4 + ^{10}C_7 \times 0.4^{8} \times (0.6)^3 +^{10}C_8 \times 0.4^{9} \times (0.6)^2$

$P(x = 11) = 0.1737$ ---- approximated

Hence, the probability he sells the candies on or before the 8th house is 0.1737

Read more about probabilities at:

brainly.com/question/251701