1. On a radio program, tickets to a concert will be given to...

1. On a radio program, tickets to a concert will be given to the fourth caller that correctly identifies a song being played by the radio station. The probability that a caller correctly identifies the song is 0.7. Find the probability that (a) the tickets will be given to the ninth caller. (b) there will be at least six callers. (c) Find the expected number of calls. Ans. (a) 0.03627 (b) 0.47178 (c) 5.714 2. A continuous random variable has density function f(y) = ky^2, 0 ? y ? 1. Find (a) k, (b) E(Y), (c) V(Y) and (d) P(Y ? 0.5). Ans. (a) 3 (b) 3/4 (c) 3/80 (d) 7/8 3. A continuous random variable Y has moment generating function m(t) = e^{50t+25t^2/2}. Find (a) P(40 ? Y ? 45) (b) a value b such that P(Y ? b) = 0.975. Ans. (a) 0.1359 (b) 59.8

Transcript

00:01 So in this question we're told, first of all, that tickets to a concert will be given to the fourth caller that correctly identifies a song.

00:09 And the probability of correctly identifying the song is 0 .7.

00:14 So that x be the number of the fourth correct caller.

00:25 And the probability x equals x is the probability that there are three correct calls in the first x minus one.

00:34 So that's x minus 1 choose 3 times 0 .7 cubed times 0 .3 to the x minus 4.

00:44 And then we need to multiply that by the probability that the x -c caller identifies it correctly.

00:49 So that's multiplied by 0 .7.

00:53 So x minus 1 choose 3, 0 .7 to the 4, 0 .3 to the x minus 4 .0 .3 to the x minus 4.

01:03 And this is valid for x in 4, 5, 6 and upwards.

01:14 So what's the probability that it will be given to the 9th caller? so that is going to be 8, choose 3 times 0 .7 to the 4 times 0 .3 to the power of 9 minus 4, which is 5, which is 0 .03627.

01:41 Part b, what's the probability there will be at least six callers? well, the probability x is greater than or equal to six, is one minus the probability x is less than or equal to five.

01:51 But since x can only be four, five and upwards, this is one minus the probability x equals four minus the probability x equals five.

02:00 So one minus three choose three times 0 .7 to the four, and then 0 .3 doesn't come in at all.

02:11 Minus 3 choose...

02:13 Sorry, minus 4 choose 3 times 0 .7 to the 4 times 0 .3 to the power of 1.

02:25 And this gives me 0 .47178.

02:31 And then the probability...

02:33 We want the expected number of calls.

02:35 Well, the expected value of x is the sum from x equals 4 to infinity, x x minus 1 choose 3 0 .7 to the 4 0 .3 to the x minus 4 which is the sum from x equals 4 to infinity x factorial over x minus 4 factorial 3 factorial because i've put the x into the x minus 1 factorial to form x factorial 0 .7 to the 4 0 .3 to the x minus 4.

03:12 So what i'm going to do is i'm going to say that y is equal to x plus 1.

03:20 Then the expected value of x is the sum from y equals 5 to infinity.

03:26 Y minus 1 choose so y minus 1 factorial divided by y minus 1 minus 4 so that's y minus 5 factorial.

03:43 And then we've got this 3 factorial down here.

03:51 But actually what i can do is i can write the y minus 5 as y minus 1 minus 4.

03:58 Then the 3 factorial i can write as 4 factorial but put a 4 up here.

04:04 Then we get 0 .7 to the power of 5 divided by 0 .7 and 0 .3 to the power of y minus 5 because x is y minus 1.

04:24 So what we get then is 4 over 0 .7 times the sum from y equals 5 to infinity, y minus 1 choose 4, 0 .7 to the 5, 0 .3 to the y minus 5.

04:42 And what you can see here is that this has the same distribution as before, but with y instead of x and 4 instead of 3.

04:52 And since we're summing from 5 now instead of 4, this is just the same distribution but shifted up by 1, and that means that its sum must be equal to 1 since it's a pmf...

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