<p>Any questions you know answers to...big help. Thankya. Show all the calculations pleeeease. :)</p>
<p>1) What's the probability of getting exactly 2 threes in 5 rolls of dice?</p>
<p>2) Justine has 5 different CDs and can place up to 3 in her cd changer.
(a) How many different listening sequences are there if she puts 3 cds in the player?
(b) How many different listening sequences are there if she puts in either 1, 2, or 3 of them at a time?</p>
<p>3) A b-ball player has made 80 per cent of her free throws over a long season. At the end of the tournament she attempts 5 free throws and only makes 2. Her fans say she choked under pressure but 2/5 of her free throws could be a result of random variation or chance. Viewing the problem as a binomial experiment, calculate the probability that she will miss 3 or more out of 5 free throws.</p>
<p>4) If there are 5 people in a room what's the probability at least 2 were born on the same day of the week?</p>
<p>No tengo mucho tiempo ahora, pero puedo ayudar un poco.</p>
<p>1) So, the probability of getting a 3= (1/6). You need exactly 2 "3"s. Now, you must arrange all the differant combinations where 3s could appear. Easiest to do with a list, IMO. *=any number</p>
<p>22**<br>
2<em>2</em>
2<strong>2*
2</strong><em>2
22</em>
<em>2</em>2*
<em>2</em><em>2
</em>22
*<em>2</em>2</p>
<p>Okay-- and there are a total of 30 combinations (6*5)
So basically, 9/30.. = 3/10 = .3</p>
<p>I'm not sure if that is right, but methinks so.</p>
<p>1) What's the probability of getting exactly 2 threes in 5 rolls of dice?
A: a three is one of six possible answers on a die, so the probability would be:
(1/6)(1/6)(5/6)(5/6)(5/6)= the probability of getting exactly 2 threes in 5 rolls of the dice</p>
<p>2) Justine has 5 different CDs and can place up to 3 in her cd changer.
(a) How many different listening sequences are there if she puts 3 cds in the player?
A: 3<em>2</em>1= 6 (the number of different listening sequences)</p>
<p>(b) How many different listening sequences are there if she puts in either 1, 2, or 3 of them at a time?
A: 1+2+ (3<em>2</em>1)= 9 (the number of different listening sequences) </p>
<p>3) A b-ball player has made 80 per cent of her free throws over a long season. At the end of the tournament she attempts 5 free throws and only makes 2. Her fans say she choked under pressure but 2/5 of her free throws could be a result of random variation or chance. Viewing the problem as a binomial experiment, calculate the probability that she will miss 3 or more out of 5 free throws.</p>
<p>4) If there are 5 people in a room what's the probability at least 2 were born on the same day of the week?</p>
<p>Oops, accidentally hit enter.... here's the whole thing</p>
<p>1) What's the probability of getting exactly 2 threes in 5 rolls of dice?</p>
<pre><code> A: a three is one of six possible answers on a die, so the probability would be:
(1/6)(1/6)(5/6)(5/6)(5/6)= the probability of getting exactly 2 threes in 5 rolls of the dice
</code></pre>
<p>2) Justine has 5 different CDs and can place up to 3 in her cd changer.
(a) How many different listening sequences are there if she puts 3 cds in the player?</p>
<pre><code> A: 3*2*1= 6 (the number of different listening sequences)
</code></pre>
<p>(b) How many different listening sequences are there if she puts in either 1, 2, or 3 of them at a time?</p>
<pre><code> A: 1+2+ (3*2*1)= 9 (the number of different listening sequences)
</code></pre>
<p>3) A b-ball player has made 80 per cent of her free throws over a long season. At the end of the tournament she attempts 5 free throws and only makes 2. Her fans say she choked under pressure but 2/5 of her free throws could be a result of random variation or chance. Viewing the problem as a binomial experiment, calculate the probability that she will miss 3 or more out of 5 free throws.</p>
<pre><code> A: on calculator: binompdf(5,.2,3)= the probability that she will miss 3 or more....
</code></pre>
<p>4) If there are 5 people in a room what's the probability at least 2 were born on the same day of the week?
A: I'm not really sure about this one, but I think its
(1/7)(1/7)(6/7)(6/7)(6/7)= the probability that at least 2 were born on the same day of the week</p>
<p>Number 2 should be
There are 5<em>4</em>3 cds in order so a total of 60 combinations.
If she puts one in, there are 5 different sequences
2 cds is 5<em>4=20 sequences
3 cds is 5</em>4*3=60
So 1or2or3 = 1+2+3 = 5 + 20 +60= 85 Ways.</p>
<p>its Pr(5,3) of 5!/(5-3)!</p>
<p>NUMBERS 1 and 3 I have not done before, but I think they are something like this
Would number one be
C(5,2)(1/6)^2 * (5/6)^3 / 6^5
3)
C(5,3)(.2)^3(.8)^2 + C(5,4)(.2)^4(.8) + C(5,5)(.2)^5 divided by 5^5(I am unsure on what to divide by here)</p>
<p>I am sure on 4 its like the birthday problem
4) it is 1-prob that no1 is born on the same day
So 1 - 7<em>6</em>5<em>4</em>3 / 7^5 = 85% chance The 7fact thing is just saying person A can have 7 days as a bday, for person B to have a different birthday, he has 6 possible birthdays and so force. Then you divide by 7^5 possible birthdays. THis is the prob that no1 has the same bday. Subtract that from one to get that atleast one has the same.</p>
<p>Smartblonde's explanation to 1 I know is wrong. While that is one possibility, it is also necessary to look at the other places where twos could end up, not just the first 2 dice. If the problem asked what the probability of getting exactly 2 twos in the first two dice was, her answer would be right.</p>
<p>Number one is this, number three is done the same</p>
<p>C(5,2)(1/6)^2*(5/6)^3 divided by the sum of j=0 to5 C(5,j) (1/6)^j * (5/6)^(5-j)<br>
If anyone knows a simpler formula than the summation, then post, but that will in fact give you the right answer.</p>
<p>Number 3 is done the same way, I dont feel like doing the summation</p>