An airline offers a survey for its passengers to complete after every flight. Each time a passenger completes the survey, there is a 2\%2%2, percent chance they will win a discounted price on their next flight. Assume that winners are selected at random, and the results of the surveys are independent. Zaylee has numerous trips planned with this airline, and she'll always complete each survey in hopes of winning. Let NNN be the number of surveys Zaylee completes until she wins for the first time. Find the probability that it takes Zaylee 333 surveys or less to win for the first time. You may round your answer to the nearest hundredth.

The probability that it takes Zaylee 3 surveys or less to win for the first time = 0.06

Step-by-step explanation:

This is a negative binomial distribution problem. A negative binomial distribution problem describes a sequence of trials, each of which can have two outcomes (success or failure). We continue the trials indefinitely until we get r succeses.

The probability of y trials before obtaining r successes is given as

P(y) = ʸ⁻¹Cᵣ₋₁ pʳ qʸ⁻ʳ (for all y ≥ r)

p = probability of success

q = probability of failure

r = number of successes required

y = number of trials required

For this question

p = 2% = 0.02

q = 1 - p = 1 - 0.02 = 0.98

r = 1

y = 3 surveys or less = ≤ 3

P(y ≤ 3) = P(y=1) + P(y=2) + P(y=3)

For P(y=1)

r = 1, y = 1

P(y=1) = 1 × (0.02)¹ (0.98)¹⁻¹ = 1 × 0.02 × 0.98⁰

= 0.02

For P(y=2)

r = 1, y = 2

P(y=2) = ¹C₀ (0.02)¹ (0.98)²⁻¹

= 1 × 0.02 × 0.98 = 0.0196

For P(y=3)

r = 1, y = 3

P(y=3) = ²C₀ (0.02)¹ (0.98)³⁻¹

= 1 × 0.02¹ × 0.98² = 0.019208

P(y ≤ 3) = P(y=1) + P(y=2) + P(y=3)

= 0.02 + 0.0196 + 0.019208 = 0.058808 = 0.06 to the nearest hundredth

Hope this Helps!!!

a c

step-by-step explanation: