SOLVED:A man tosses a fair coin 10 times. Find the probability that he will have: (i) heads on the first five tosses and tails on the next five tosses, (ii) heads on tosses 1,3,5,7,9 and tails on 2,4,

Video Transcript

in this question were given that a coin is tossed 10 times and the coin is fair. And we want to answer the probability interrogate from part one to character five before we do that let ‘s specify heat content to be the event where we get ahead in the convulse and TBD consequence we get a tail in the discard. now we know that probability of getting ahead is half because the coins fifth wheel and probability of getting a there is besides half. immediately the number of trials is 10 and the probability of getting a achiever which are defined to be getting a collision to be half and that one-half is changeless. So we actually have a binomial distribution here. Let X to be the distribution of successes which is getting hits. thus adam actually follows a binomial distribution trial is 10 and probably or successes half. So this is the formula for binomial distribution and choose X. nowadays our m is actually 10. So I ‘m merely gon na put 10 um probability of success is hard to the power X. And property of bankruptcy is one minus half and that ‘s half that ‘s carrying a stern 10 subtraction X. Okay, now in partially one we want to find the probability of heads in the first five pass and buttocks in the future five towards that means we want this. Okay so this one we ‘re not using by normal because it ‘s fixed that the first gear five forces ahead and the next five tosses are tails. So probability of getting our first forefront is half lapp thing. 2nd promontory was besides behalf as they ‘re autonomous so you can see it will be half all the way. There is besides a probability of half all the way until you get tom ‘s help. So that is half of the power of 10. And therefore B 1/10- four. Next we want to find the probability of getting the heads on discard 13579 and tails on 2468 10. So it will be much mating like this. indeed again getting ahead in the first course is half getting a chase in the second tosses half therefore on and so away. You ‘ll get half the region 10 again and It will be 1/10-4. now the separate probably you ‘re getting five tails. five heads and five tails because this promontory can appear in um not just the first five toss it can appear anywhere in between flip number 1 to 10. so for this we will be using the binomial distribution. So this is actually asking for p. ten equals 25. So let ‘s good sub in. It goes to 10, choose five Half to the exponent of five And one-half to the baron of 10 -5. and so this will be 63 over 256. Okay, next separate four We want to find probability of at least five hits. That means we want to find probability of X. Greater equals to five. now this if you ‘re not using a calculator to calculate the statistics. If we do manually It will be p. x. equals 2 5 Plus PX equals 2. 6 Plus PX equals 2 7. All the direction to plus P. X. Equals to thank. Okay so Ps goes to five. We already know. Is this 63 out of 256 x equals to six will be 10-6 Half to the baron six and then half the mighty 10 to 7 for PS 47 half to half to about seven half to about three. All the way to the last one would be then Tuesday then this one dainty daniel half to the power 10.5 the exponent zero. thus calculating this, If you ‘re bang-up to a calculator you get 319 out of 512. o in part five we want to find the probability of not more than five heads. not more than five heads would be P. X. less than equals to five. yea but so I want to use P 1 -4. ten. Greater than five. So that means it will be one P. x equals 26 ten equals 2 7. All the way to Mexico ‘s 2 10. So it will be one As it goes to six will be 10-6 Half to the power six half to the power of four Plus 10- seven. Half the baron seven have to the power of three. All the way to the final one would be, then choose 10. Have to rip out, then Have to Power nothing. then if you were to compute this, your answer will be 319.512.

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