Question

suppose dan loses 28% of all bingo games. (a) what is the probability that dan loses two bingo games in a row? (b) what is the probability that dan loses six bingo games in a row? (c) when events are independent, their complements are independent as well. use this result to determine the probability that dan loses six bingo games in a row, but does not lose seven in a row. (a) the probability that dan loses two bingo games in a row is . (round to four decimal places as needed.)

Explanation:

Step1: Identify the probability of losing a single game

The probability of losing a single bingo - game, $p = 0.28$.

Step2: Use the multiplication rule for independent events

For independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$. When losing two games in a row, the probability is $P = p\times p$.
$P=(0.28)\times(0.28)=0.0784$

Step3: Probability of losing six games in a row

Using the multiplication rule for independent events, if we want to find the probability of losing six games in a row, $P = p^6$.
$P=(0.28)^6=0.28\times0.28\times0.28\times0.28\times0.28\times0.28\approx0.000482$

Step4: Probability of losing six but not seven games in a row

The probability of losing six games in a row is $p^6$, and the probability of winning the seventh game is $1 - p$. So the probability of losing six games in a row but not seven is $p^6\times(1 - p)$.
$P=(0.28)^6\times(1 - 0.28)=(0.28)^6\times0.72\approx0.000482\times0.72\approx0.000347$

Answer:

(a) $0.0784$
(b) $0.0005$ (rounded to four decimal places)
(c) $0.0003$ (rounded to four decimal places)