9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - What is the minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.95? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined If the coin is tossed once, there will be 2^1 = 2 choices i.e. =\\H, T\\ If the coin is tossed twice, there will be 2^2 = 4 choices i.e. =\\HH, HT, TH, TT\\ If the coin is tossed thrice, there will be 2^3 = 8 choices i.e. =\\HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\\ ...and so on.

Instead of calculating \"at <=ast one head\" directly, we'll use the comp<=ment: P(at <=ast one head) = 1 - P(all tails) There will only be one case of \"all tails\" (as you see in the 3 examp<=s above). We want the probability of \"all tails\" to be < 0.05.

Let n = number of coin tosses n = 3: Probability of all tails =18 = 0.125 n = 4: Probability of all tails =116 = 0.0625 n = 5: Probability of all tails =132 0.03125 Since 0.03125 < 0.05, n=5 works!

The minimum number of tosses is 5."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Modern Math,Probability"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - What is the minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.95? | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined If the coin is tossed once, there will be 2^1 = 2 choices i.e. =\\H, T\\ If the coin is tossed twice, there will be 2^2 = 4 choices i.e. =\\HH, HT, TH, TT\\ If the coin is tossed thrice, there will be 2^3 = 8 choices i.e. =\\HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\\ ...and so on.

Let n = number of coin tosses n = 3: Probability of all tails =18 = 0.125 n = 4: Probability of all tails =116 = 0.0625 n = 5: Probability of all tails =132 0.03125 Since 0.03125 < 0.05, n=5 works!

The minimum number of tosses is 5."}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/29"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | JIPMAT Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - What is the minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.95? | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined If the coin is tossed once, there will be 2^1 = 2 choices i.e. =\\H, T\\ If the coin is tossed twice, there will be 2^2 = 4 choices i.e. =\\HH, HT, TH, TT\\ If the coin is tossed thrice, there will be 2^3 = 8 choices i.e. =\\HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\\ ...and so on.

Let n = number of coin tosses n = 3: Probability of all tails =18 = 0.125 n = 4: Probability of all tails =116 = 0.0625 n = 5: Probability of all tails =132 0.03125 Since 0.03125 < 0.05, n=5 works!

The minimum number of tosses is 5."}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21"],"error":null,"digest":"$undefined"}

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