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2 years ago

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In the drawing, six out of every 10 tickets are winning tickets. Of the winning tickets, 1 out of every three awards is a larger

prize. What is the probability that a ticket that is randomly chosen will award a larger prize?

2 answers:

mafiozo [28]2 years ago

6 0

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Given : Six out of every 10 tickets are winning tickets.

Of the winning tickets, 1 out of every three awards is a larger prize.

Solution :

Since 10 tickets are drawn

Out of 10 , 6 tickets are wining

All six will get awards

Out of three awards 1 is a larger pize and total there are six awards

So, out of six awards there are 2 larger awards

Thus out of 10 tickets there are 2 larger awards

⇒ $\frac{2}{10} = \frac{1}{5}$

Thus, The probability that a ticket that is randomly chosen will award a larger prize is

$\frac{1}{5}$

omeli [17]2 years ago

5 0

1/5 is the answer hope it helps

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2 years ago

Find the values of x and y that satisfy the equation.<br><br> 4x+2i=8+yi<br> x= <br> and y=

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2 years ago

Each airline passenger and his or her luggage must be checked to determine whether he or she is carrying weapons on to the airpl

Dafna1 [17]

Answer:

a

$P_k = 0.83$

b

$N_{\mu} \approx 4 \ passengers$

c

$T_{\lambda} = 0.5 \ minutes$

Step-by-step explanation:

From the question we are told that

The average number of passengers that arrive per minute is $\lambda = 10$

The average number of check that can be carried out in one minute is $\mu= 12$

Generally the probability that a passenger will have to wait before being checked for weapons is mathematically represented as

$P_k = \frac{\lambda }{ \mu }$

=> $P_k = \frac{10 }{ 12}$

=> $P_k = 0.83$

Generally the number of passengers are waiting in line to enter the checkpoint is mathematically represented as

$N_{\mu} = \frac{\lambda^2}{\mu (\mu -\lambda) }$

=> $N_{\mu} = \frac{10^2}{12 (12 -10) }$

=> $N_{\mu} \approx 4 \ passengers$

Generally the average time a passenger spend at the checkpoint is mathematically represented as

$T_{\lambda} = \frac{ \frac{\lambda}{(\mu - \lambda)} }{ \lambda}$

=> $T_{\lambda} = \frac{ \frac{ 10}{(12 - 10)} }{10}$

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5 0

1 year ago

An annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. If the occupants o

Sunny_sXe [5.5K]

Answer:

Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.

Step-by-step explanation:

We are given that an annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. The occupants of half the residences object to being annexed.

Also, a random sample of 10 residents is taken.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 10 residents

r = number of success = at least 3

p = probability of success which in our question is probability that

residents favor the annexation suit, which is calculated as below;

p = $\frac{\text{Number of residents favoring the annexation suit }}{\text{Total number of residents considered } }$ = $\frac{600}{1200}$ = 0.50

<em>LET X = Number of residents favoring the annexation suit</em>

So, it means X ~ $Binom(n=10, p=0.50)$

Now, Probability that in a random sample of 10 at least 3 favor the annexation suit is given by = P(X $\geq$ 3)

P(X $\geq$ 3) = 1 - P(X < 3) = 1 - P(X $\leq$ 2)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

= $1- [\binom{10}{0}\times 0.50^{0} \times (1-0.50)^{10-0} + \binom{10}{1}\times 0.50^{1} \times (1-0.50)^{10-1} +\binom{10}{2}\times 0.50^{2} \times (1-0.50)^{10-2}]$

= $1-[ 1 \times 1 \times 0.50^{10}+10 \times 0.50^{1} \times 0.50^{9}+45 \times 0.50^{2} \times 0.50^{8}]$

= $1-[ 0.50^{10}+10 \times 0.50^{10}+45 \times 0.50^{10}]$

= $1-0.50^{10}[ 1+10 +45 ]$ = $1-0.50^{10} \times 56$

= 0.9453

Therefore, Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.

8 0

2 years ago