Question

The probability of buying a movie ticket with a popcorn coupon is 0.608. If you buy 10 movie tickets, what is the probability that 3 or more of the tickets have popcorn coupons? (Round your answer to 3 decimal places if necessary.)

Answer 1

Answer:

0.989

Step-by-step explanation:

The probability of buying a movie ticket with a popcorn coupon is p = 0.608, so the probability of buying a movie ticket without a popcorn coupon is q = 1 - 0.608 = 0.392.

Find the following probabilities;

• the probability that 0 tickets have popcorn coupons $P_0=C^{10}_0p^0q^{10-0}=(0.392)^10\approx 0.000086$
• the probability that exactly one ticket has a popcorn coupon $P_1=C^{10}_1p^1q^{10-1}=10\cdot 0.608\cdot 0.392^9\approx 0.001329$
• the probability that ecatly 2 tickets have a popcorn coupons $P_2=C^{10}_2p^2q^{10-2}=45\cdot 0.608^2\cdot 0.392^8\approx 0.009275$

Hence, the probability that 3 or more of the tickets have popcorn coupons is

$P=1-P_0-P_1-P_2=1-0.000086-0.001329-0.009275=0.98931\approx 0.989$

Answer 2

Answer:

P(3 or more of the tickets have popcorn coupons) = 0.989

Step-by-step explanation:

The probability of buying a movie ticket with a popcorn coupon is 0.608

p = 0.608

The probability of buying a movie ticket without a popcorn coupon is q

q = 1 - p = 1 - 0.608 = 0.392

If you buy 10 movie tickets, n = 10

The probability that 3 or more of the tickets have popcorn coupons.

X = 3,4,5,6,7,8,9,10

Using Binomial distribution,

p = Success , q = failure, n = number of trial , r = X

$P(X=r)=^nC_rp^rq^{n-r}$

For X = 3,4,5,6,7,8,9,10

Let E be probability 3 or more of the tickets have popcorn coupons.

$P(E)=^{10}C_3(0.608)^3(0.392)^7+^{10}C_4(0.608)^4(0.392)^6+^{10}C_5(0.608)^5(0.392)^5+^{10}C_6(0.608)^6(0.392)^4+^{10}C_7(0.608)^7(0.392)^3+^{10}C_8(0.608)^8(0.392)^2+^{10}C_9(0.608)^9(0.392)^1+^{10}C_{10}(0.608)^{10}(0.392)^0$

$P(E)=0.9893$

Now round to 3 decimal place.

P(3 or more of the tickets have popcorn coupons) = 0.989