Five coins are tossed 3200 times. What is the probability of getting 5 heads exactly two times?

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Let's understand this binomial probability problem. We have five coins tossed 3200 times, and we want to find the probability of getting exactly 5 heads in exactly two of those tosses. This is a classic binomial distribution problem where success is getting all 5 heads in one toss. Now let's identify the parameters for our binomial distribution. We have n equals 3200 trials, we want k equals 2 successes, and the probability of success p is 1 over 32, since getting 5 heads has probability one-half to the fifth power. The probability of failure q is 31 over 32. We'll use the binomial formula to calculate the final probability. Let's calculate the combination C of 3200 choose 2. This equals 3200 factorial divided by 2 factorial times 3198 factorial. We can simplify this to 3200 times 3199 divided by 2, which gives us 10,236,800 divided by 2, resulting in 5,118,400. This represents the number of ways to choose exactly 2 tosses out of 3200 tosses. Now we apply the complete binomial formula. We multiply 5,118,400 by one over 32 squared, which is one over 1024, and by 31 over 32 to the power of 3198. Simplifying, we get 5,118,400 divided by 1024, which equals exactly 5000, times 31 over 32 to the power of 3198. This is our final answer for the probability. In conclusion, the probability of getting exactly 5 heads in exactly 2 out of 3200 tosses is 5000 times 31 over 32 to the power of 3198. This is an extremely small probability because the term 31 over 32 raised to the 3198th power is incredibly tiny, reflecting the rarity of this specific outcome in a binomial distribution.